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Algebra Quiz
Practice Solve Quadratics By Multiple Methods in Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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Solve by factoring: x2+7x+12=0
This quiz focuses on Solve Quadratics By Multiple Methods, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
Solve by factoring: x2+7x+12=0
Explanation: This question tests your ability to solve quadratic equations using factoring, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor x² + 7x + 12 to get (x + 3)(x + 4) = 0. Using the Zero Product Property, either (x + 3) = 0 or (x + 4) = 0. Solving each: x + 3 = 0 gives x = -3, and x + 4 = 0 gives x = -4. These are the two solutions! Choice B is correct because it properly applies the factoring method and identifies both solutions: x = -3 and x = -4. Choice A gives positive solutions x = 3, 4, but when we have (x + 3)(x + 4) = 0, solving x + 3 = 0 gives x = -3 (not x = 3). When you solve x + 3 = 0, you subtract 3 from both sides, giving the negative value. Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!
Solve by factoring: x2+9x+20=0
Explanation: This question tests your ability to solve quadratic equations by factoring, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor x² + 9x + 20 to get (x + 4)(x + 5) = 0. Using the Zero Product Property, either (x + 4) = 0 or (x + 5) = 0. Solving each: x = -4 and x = -5. These are the two solutions! Choice C is correct because it properly applies the method and includes both solutions with correct arithmetic. Well done! Choice A solves the Zero Product Property incorrectly: from (x + 4) = 0, we get x = -4 (not x = 4). When you solve x + 4 = 0, you subtract 4 from both sides, giving the negative value. Here's how to choose your method: (1) If it's x² = number, use inspection or square roots—super fast! (2) If it factors easily (you can spot the factors quickly), use factoring. (3) If it doesn't factor nicely or you need complex solutions, use the quadratic formula. With practice, you'll recognize which method fits each equation instantly! Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!
Solve: x2−12x+36=0
Explanation: This question tests your ability to solve quadratic equations using multiple methods, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor x² - 12x + 36 to get (x - 6)(x - 6) = 0, or (x - 6)² = 0. Using the Zero Product Property, x - 6 = 0, giving the repeated solution x = 6. This is a perfect square trinomial with a double root! Choice A is correct because it properly identifies the repeated root. Well done! Choice C has two different solutions when it's actually a repeated root; quadratics like this have two solutions counting multiplicity, but they are the same value. Here's how to choose your method: (1) If it's x² = number, use inspection or square roots—super fast! (2) If it factors easily (you can spot the factors quickly), use factoring. (3) If it doesn't factor nicely or you need complex solutions, use the quadratic formula. With practice, you'll recognize which method fits each equation instantly! The ± symbol is your reminder to find both solutions, but in cases like this, they coincide into one. Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!
Solve x2+2x+5=0. Express complex solutions in a±bi form.
Explanation: This question tests your ability to solve quadratic equations using the quadratic formula, including complex solutions, which is an essential skill in Algebra 1. The quadratic formula x = (-b ± √(b² - 4ac))/(2a) works for ANY quadratic equation ax² + bx + c = 0—it's your reliable backup method when other approaches don't work easily, and it also reveals when you have complex solutions (when the part under the square root, b² - 4ac, is negative). When we calculate the discriminant b² - 4ac = 4 - 20 = -16, we get a negative number under the square root. This means the solutions are complex (involving i, where i = √(-1)). Working through the quadratic formula: x = (-2 ± √(-16))/2 = (-2 ± 4i)/2 = -1 ± 2i. These complex solutions come in a conjugate pair! Choice A is correct because it properly expresses complex solutions correctly as a ± bi. Well done! Choice D forgets to include i in the complex solution, giving -1 ± 2 instead of -1 ± 2i. When the discriminant is negative, the square root involves i = √(-1), making the solutions complex. When you get a negative discriminant (b² - 4ac < 0), don't panic! It just means your solutions involve i. Calculate √(-discriminant), then put an i with it: √(-16) = 4i. Your solutions will be complex numbers in the form a + bi and a - bi. Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!
Solve x2+2x+2=0. Express complex solutions in a±bi form.
Explanation: This question tests your ability to solve quadratic equations using the quadratic formula for complex solutions, which is an essential skill in Algebra 1. The quadratic formula x = (-b ± √(b² - 4ac))/(2a) works for ANY quadratic equation ax² + bx + c = 0—it's your reliable backup method when other approaches don't work easily, and it also reveals when you have complex solutions (when the part under the square root, b² - 4ac, is negative). When we calculate the discriminant b² - 4ac = 4 - 8 = -4, we get a negative number under the square root. This means the solutions are complex (involving i, where i = √(-1)). Working through the quadratic formula: x = (-2 ± √(-4))/2 = (-2 ± 2i)/2 = -1 ± i. These complex solutions come in a conjugate pair! Choice B is correct because it properly expresses complex solutions correctly as a ± bi. Well done! Choice D forgets to include i in the complex solution, giving -1 ± 2 instead of -1 ± i. When the discriminant is negative, the square root involves i = √(-1), making the solutions complex. When you get a negative discriminant (b² - 4ac < 0), don't panic! It just means your solutions involve i. Calculate √(-discriminant), then put an i with it: √(-4) = 2i. Your solutions will be complex numbers in the form a + bi and a - bi. For the quadratic formula, here's a trick: write out 'x = (-b ± √(b² - 4ac))/(2a)' before you start, then carefully substitute a, b, and c from your equation. If you try to do it in your head, it's easy to drop a negative sign or mix up the numbers.
Solve by taking square roots: (x−4)2=25. Find all solutions.
Explanation: This question tests your ability to solve quadratic equations using taking square roots, which is an essential skill in Algebra 1. When solving by taking square roots from an equation like (x - p)² = k, take the square root of both sides to get x - p = ±√k, then solve for x: x = p ± √k. The ± symbol means you get two solutions: p + √k and p - √k. Starting with (x - 4)² = 25, we take the square root of both sides: x - 4 = ±5. Now we solve for x by adding 4 to both sides: x = 4 ± 5. This gives us two solutions: x = 4 + 5 = 9 and x = 4 - 5 = -1. The ± splits into two separate calculations! Choice C is correct because it properly expresses the solutions as x = 4 ± 5, which represents both x = 9 and x = -1. Choice A has the right final answers but doesn't show the ± form, while Choice B makes an error by writing ±25 instead of ±5 (we take the square root of 25, which is 5, not 25 itself). Quick check: after solving, substitute your answers back into the original equation. (9-4)² = 5² = 25 ✓ and (-1-4)² = (-5)² = 25 ✓. If you get the original right side, great! This catch-your-own-mistakes habit is one of the best math skills you can develop!
A ball’s height (in feet) is modeled by h(t)=−16t2+32t. When does it hit the ground (solve h(t)=0)?
Explanation: This question tests your ability to solve quadratic equations by factoring in a real-world context, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor -16t² + 32t = 0 to get -16t(t - 2) = 0. Using the Zero Product Property, either -16t = 0 or (t - 2) = 0. Solving each: t = 0 and t = 2. These are the two solutions, representing when the ball is thrown (t=0) and when it hits the ground (t=2)! Choice B is correct because it properly applies the method and includes both solutions with correct arithmetic. Well done! Choice A solves the Zero Product Property incorrectly: from (t - 2) = 0, we get t = 2 (not t = -2). When you solve t - 2 = 0, you add 2 to both sides, giving the positive value. Here's how to choose your method: (1) If it's x² = number, use inspection or square roots—super fast! (2) If it factors easily (you can spot the factors quickly), use factoring. (3) If it doesn't factor nicely or you need complex solutions, use the quadratic formula. With practice, you'll recognize which method fits each equation instantly! Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!
Solve by factoring: x2−11x+24=0
Explanation: This question tests your ability to solve quadratic equations using factoring, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor x² - 11x + 24 to get (x - 3)(x - 8) = 0. Using the Zero Product Property, either (x - 3) = 0 or (x - 8) = 0. Solving each: x - 3 = 0 gives x = 3, and x - 8 = 0 gives x = 8. These are the two solutions! Choice A is correct because it properly applies the factoring method and identifies both solutions: x = 3 and x = 8. Choice B has sign errors—when we factor x² - 11x + 24, we need two numbers that multiply to +24 and add to -11, which are -3 and -8, giving us (x - 3)(x - 8) = 0, not (x + 3)(x + 8) = 0. Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!
Solve: x2=64
Explanation: This question tests your ability to solve quadratic equations using inspection or taking square roots, which is an essential skill in Algebra 1. When a quadratic is in the simple form x2=k, we can solve by inspection (just recognizing the answer) or by taking the square root of both sides, remembering that we need both the positive and negative square roots: x=±k. Looking at x2=64, we recognize that 64 is a perfect square, 82. Taking the square root of both sides: x=±8. This means x=8 or x=−8. Both work because 82=64 AND (−8)2=64. Always remember that ± symbol—it gives us both solutions! Choice B is correct because it properly includes both solutions with correct arithmetic. Well done! Choice A forgets the negative root—a super common mistake! When we take the square root, we always get ± (both positive and negative), so x2=64 gives x=±8, which means both x=8 AND x=−8. Here's how to choose your method: (1) If it's x2= number, use inspection or square roots—super fast! (2) If it factors easily (you can spot the factors quickly), use factoring. (3) If it doesn't factor nicely or you need complex solutions, use the quadratic formula. With practice, you'll recognize which method fits each equation instantly! Quick check: after solving, substitute your answers back into the original equation. If you get 0=0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!
Using the quadratic formula, solve x2+5x+6=0.
Explanation: This question tests your ability to solve quadratic equations using the quadratic formula, which is an essential skill in Algebra 1. The quadratic formula x = (-b ± √(b² - 4ac))/(2a) works for ANY quadratic equation ax² + bx + c = 0—it's your reliable backup method when other approaches don't work easily, and it also reveals when you have complex solutions (when the part under the square root, b² - 4ac, is negative). Using the quadratic formula with a = 1, b = 5, c = 6: x = (-5 ± √(25 - 24))/2 = (-5 ± √1)/2 = (-5 ± 1)/2. This simplifies to x = (-5 + 1)/2 = -2 and x = (-5 - 1)/2 = -3. The ± in the formula is what gives us two solutions! Choice A is correct because it properly applies the method and includes both solutions with correct arithmetic. Well done! Choice D doesn't simplify the radical: √1 can be simplified to 1. When using the quadratic formula, always simplify your radicals and fractions at the end! For the quadratic formula, here's a trick: write out 'x = (-b ± √(b² - 4ac))/(2a)' before you start, then carefully substitute a, b, and c from your equation. If you try to do it in your head, it's easy to drop a negative sign or mix up the numbers. Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!
Solve by taking square roots: (x−4)2=25
Explanation: This question tests your ability to solve quadratic equations using taking square roots, which is an essential skill in Algebra 1. When solving by taking square roots from an equation like (x−p)2=k, take the square root of both sides to get x−p=±k, then solve for x: x=p±k. The ± symbol means you get two solutions: p+k and p−k. Starting with (x−4)2=25, we take the square root of both sides: x−4=±5. Now we solve for x by adding 4 to both sides: x=4±5. This gives us two solutions: x=4+5=9 and x=4−5=−1. The ± splits into two separate calculations! Choice A is correct because it properly applies the method and includes both solutions: x=9 and x=−1. Choice B gives x=±5, which would be the answer if we had x2=25, but here we have (x−4)2=25, so we need to account for the -4 inside the parentheses. The ± symbol is your reminder to find both solutions: x=4±5 means two answers (4+5=9 and 4−5=−1). Don't just calculate the plus—do both! This is one of the most common places students lose a solution.
A ball’s height (in feet) is modeled by h(t)=−16t2+64t. When does it hit the ground (solve h(t)=0)?
Explanation: This question tests your ability to solve quadratic equations using factoring, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor h(t) = -16t² + 64t = 0 by first factoring out the common factor of -16t: -16t(t - 4) = 0. Using the Zero Product Property, either -16t = 0 or (t - 4) = 0. Solving each: -16t = 0 gives t = 0, and t - 4 = 0 gives t = 4. These are the two solutions! Choice A is correct because it properly identifies when the ball hits the ground: at t = 0 (the initial launch) and t = 4 (when it returns to ground level after 4 seconds). Choice D gives t = 0, 64, but 64 comes from confusing the coefficient 64t with a solution—always solve the factored equation, not just read numbers from the original! Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!
Solve by factoring: x2+7x+12=0
Explanation: This question tests your ability to solve quadratic equations using factoring, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor x² + 7x + 12 to get (x + 3)(x + 4) = 0. Using the Zero Product Property, either (x + 3) = 0 or (x + 4) = 0. Solving each: x + 3 = 0 gives x = -3, and x + 4 = 0 gives x = -4. These are the two solutions! Choice B is correct because it properly identifies both solutions as x = -3 and x = -4. Well done! Choice A makes a sign error: from (x + 3) = 0, we get x = -3 (not x = 3). When you solve x + 3 = 0, you subtract 3 from both sides, giving the negative value. Here's how to choose your method: (1) If it's x² = number, use inspection or square roots—super fast! (2) If it factors easily (you can spot the factors quickly), use factoring. (3) If it doesn't factor nicely or you need complex solutions, use the quadratic formula. With practice, you'll recognize which method fits each equation instantly!
Solve by factoring: x2−7x+12=0
Explanation: This question tests your ability to solve quadratic equations by factoring, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor x² - 7x + 12 to get (x - 3)(x - 4) = 0. Using the Zero Product Property, either (x - 3) = 0 or (x - 4) = 0. Solving each: x = 3 and x = 4. These are the two solutions! Choice B is correct because it properly applies the method and includes both solutions with correct arithmetic. Well done! Choice A solves the Zero Product Property incorrectly: from (x - 3) = 0, we get x = 3 (not x = -3). When you solve x - 3 = 0, you add 3 to both sides, giving the positive value here. Here's how to choose your method: (1) If it's x² = number, use inspection or square roots—super fast! (2) If it factors easily (you can spot the factors quickly), use factoring. (3) If it doesn't factor nicely or you need complex solutions, use the quadratic formula. With practice, you'll recognize which method fits each equation instantly! Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!
Using the quadratic formula, solve x2+6x+5=0.
Explanation: This question tests your ability to solve quadratic equations using the quadratic formula, which is an essential skill in Algebra 1. The quadratic formula x=(−b±b2−4ac)/(2a) works for ANY quadratic equation ax2+bx+c=0—it's your reliable backup method when other approaches don't work easily, and it also reveals when you have complex solutions. Using the quadratic formula with a = 1, b = 6, c = 5: x=(−6±62−4(1)(5))/(2⋅1)=(−6±36−20)/2=(−6±16)/2=(−6±4)/2. This gives us x=(−6+4)/2=−2/2=−1 and x=(−6−4)/2=−10/2=−5. The ± in the formula is what gives us two solutions! Choice A is correct because it properly applies the quadratic formula and gets both solutions: x = -1 and x = -5. Well done! Choice B has a sign error in the quadratic formula: the formula starts with -b (negative b), so with b = 6, we use -6, not +6. Those signs are tricky! For the quadratic formula, here's a trick: write out 'x = (-b \pm \sqrt{b^2 - 4ac})/(2a)' before you start, then carefully substitute a, b, and c from your equation. If you try to do it in your head, it's easy to drop a negative sign or mix up the numbers.
Solve by factoring: x2+5x+6=0
Explanation: This question tests your ability to solve quadratic equations by factoring, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor x² + 5x + 6 to get (x + 2)(x + 3) = 0. Using the Zero Product Property, either (x + 2) = 0 or (x + 3) = 0. Solving each: x = -2 and x = -3. These are the two solutions! Choice B is correct because it properly applies the method and includes both solutions with correct arithmetic. Well done! Choice A solves the Zero Product Property incorrectly: from (x + 2) = 0, we get x = -2 (not x = 2). When you solve x + 2 = 0, you subtract 2 from both sides, giving the negative value. Here's how to choose your method: (1) If it's x² = number, use inspection or square roots—super fast! (2) If it factors easily (you can spot the factors quickly), use factoring. (3) If it doesn't factor nicely or you need complex solutions, use the quadratic formula. With practice, you'll recognize which method fits each equation instantly! Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!
Solve x2−7x+12=0 by factoring and the Zero Product Property.
Explanation: This question tests your ability to solve quadratic equations by factoring, which is an essential skill in Algebra 1. For factoring method: once you have the quadratic factored as (x - p)(x - q) = 0, the Zero Product Property tells us that either x - p = 0 or x - q = 0, giving us the two solutions x = p and x = q. We factor x² - 7x + 12 to get (x - 3)(x - 4) = 0. Using the Zero Product Property, either (x - 3) = 0 or (x - 4) = 0. Solving each: x = 3 and x = 4. These are the two solutions! Choice B is correct because it properly applies the method and includes both solutions with correct arithmetic. Well done! Choice A solves the Zero Product Property incorrectly: from (x - 3) = 0, we get x = 3 (not x = -3). When you solve x - 3 = 0, you add 3 to both sides, giving the positive value. Here's how to choose your method: (1) If it's x² = number, use inspection or square roots—super fast! (2) If it factors easily (you can spot the factors quickly), use factoring. (3) If it doesn't factor nicely or you need complex solutions, use the quadratic formula. With practice, you'll recognize which method fits each equation instantly! Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!
Using the quadratic formula, solve x2−4x+8=0. Express complex solutions in a±bi form.
Explanation: This question tests your ability to solve quadratic equations using the quadratic formula, which is an essential skill in Algebra 1. The quadratic formula x = (-b ± √(b² - 4ac))/(2a) works for ANY quadratic equation ax² + bx + c = 0—it's your reliable backup method when other approaches don't work easily, and it also reveals when you have complex solutions (when the part under the square root, b² - 4ac, is negative). Using the quadratic formula with a = 1, b = -4, c = 8: x = (-(-4) ± √((-4)² - 4(1)(8)))/(2·1) = (4 ± √(16 - 32))/2 = (4 ± √(-16))/2 = (4 ± 4i)/2 = 2 ± 2i. When we calculate the discriminant b² - 4ac = 16 - 32 = -16, we get a negative number under the square root. This means the solutions are complex (involving i, where i = √(-1)). These complex solutions come in a conjugate pair! Choice A is correct because it properly expresses the complex solutions as x = 2 ± 2i, which means x = 2 + 2i and x = 2 - 2i. Choice D gives real solutions x = 2 ± 2 when the discriminant is negative. When b² - 4ac < 0, like 16 - 32 = -16, there are no real solutions—only complex ones involving i. When you get a negative discriminant (b² - 4ac < 0), don't panic! It just means your solutions involve i. Calculate √(-discriminant), then put an i with it: √(-16) = 4i. Your solutions will be complex numbers in the form a + bi and a - bi.
Using the quadratic formula, solve 2x2−4x−6=0.
Explanation: This question tests your ability to solve quadratic equations using the quadratic formula, which is an essential skill in Algebra 1. The quadratic formula x = (-b ± √(b² - 4ac))/(2a) works for ANY quadratic equation ax² + bx + c = 0—it's your reliable backup method when other approaches don't work easily, and it also reveals when you have complex solutions (when the part under the square root, b² - 4ac, is negative). Using the quadratic formula with a = 2, b = -4, c = -6: x = (-(-4) ± √((-4)² - 4(2)(-6)))/(2·2) = (4 ± √(16 + 48))/4 = (4 ± √64)/4 = (4 ± 8)/4. This gives x = (4 + 8)/4 = 12/4 = 3 and x = (4 - 8)/4 = -4/4 = -1. The ± in the formula is what gives us two solutions! Choice A is correct because it properly applies the method and includes both solutions with correct arithmetic. Well done! Choice B has a sign error in the quadratic formula: with b = -4, -b = 4 (positive), but perhaps confusing the order of solutions. Those signs are tricky! For the quadratic formula, here's a trick: write out 'x = (-b ± √(b² - 4ac))/(2a)' before you start, then carefully substitute a, b, and c from your equation. If you try to do it in your head, it's easy to drop a negative sign or mix up the numbers. Quick check: after solving, substitute your answers back into the original equation. If you get 0 = 0, great! If not, you made an error somewhere. This catch-your-own-mistakes habit is one of the best math skills you can develop!
Solve x2+2x+5=0. Express complex solutions in a±bi form.
Explanation: This question tests your ability to solve quadratic equations using the quadratic formula, which is an essential skill in Algebra 1. The quadratic formula x = (-b ± √(b² - 4ac))/(2a) works for ANY quadratic equation ax² + bx + c = 0—it's your reliable backup method when other approaches don't work easily, and it also reveals when you have complex solutions (when the part under the square root, b² - 4ac, is negative). Using the quadratic formula with a = 1, b = 2, c = 5: x = (-2 ± √(2² - 4(1)(5)))/(2·1) = (-2 ± √(4 - 20))/2 = (-2 ± √(-16))/2 = (-2 ± 4i)/2 = -1 ± 2i. When we calculate the discriminant b² - 4ac = 4 - 20 = -16, we get a negative number under the square root. This means the solutions are complex (involving i, where i = √(-1)). These complex solutions come in a conjugate pair! Choice C is correct because it properly expresses the complex solutions as x = -1 ± 2i, which means x = -1 + 2i and x = -1 - 2i. Choice A forgets to include i in the complex solution, giving -1 ± 2 instead of -1 ± 2i. When the discriminant is negative, the square root involves i = √(-1), making the solutions complex. When you get a negative discriminant (b² - 4ac < 0), don't panic! It just means your solutions involve i. Calculate √(-discriminant), then put an i with it: √(-16) = 4i. Your solutions will be complex numbers in the form a + bi and a - bi.