This question asks us to find the solutions to the quadratic equation $x^2-5x-24=0$. To solve by factoring, we need two numbers that multiply to $-24$ (not $+24$) and add to $-5$. These numbers are $-8$ and $+3$, giving us $(x-8)(x+3)=0$. Setting each factor to zero: $x-8=0$ gives $x=8$, and $x+3=0$ gives $x=-3$. The common error of using factors of $+24$ would lead to incorrect signs. When factoring quadratics with a negative constant term, remember one factor must be positive and one negative.

To solve 3x² + 2x - 8 = 0, we can use the quadratic formula: x = (-b ± √(b² - 4ac))/(2a). Here, a = 3, b = 2, and c = -8. The discriminant is b² - 4ac = 4 - 4(3)(-8) = 4 + 96 = 100. So x = (-2 ± √100)/(2·3) = (-2 ± 10)/6. This gives x = (-2 + 10)/6 = 8/6 = 4/3 or x = (-2 - 10)/6 = -12/6 = -2. The solutions are x = -2 and x = 4/3. Common errors include sign mistakes in the formula or arithmetic errors when calculating the discriminant. Always double-check your discriminant calculation and simplify fractions completely.

To find the vertex of y = 2x² - 8x + 3, we can use the vertex formula or complete the square. Using the formula, the x-coordinate of the vertex is h = -b/(2a) = -(-8)/(2(2)) = 8/4 = 2. To find the y-coordinate, substitute x = 2 back into the equation: y = 2(2)² - 8(2) + 3 = 8 - 16 + 3 = -5. Therefore, the vertex is (2, -5). A common error is stopping after finding only the x-coordinate or making sign errors when completing the square. When using the vertex formula, always substitute back to find the y-coordinate of the vertex.

We need to find when the ball reaches maximum height using h(t) = -16t² + 64t + 5. For a quadratic function in the form at² + bt + c, the maximum (when a < 0) occurs at t = -b/(2a). Here, a = -16 and b = 64, so t = -64/(2(-16)) = -64/(-32) = 2 seconds. Common mistakes include using the y-intercept (5, when t = 0) or finding the roots instead of the vertex. Since the coefficient of t² is negative, this parabola opens downward, confirming we're finding a maximum. Always use the vertex formula t = -b/(2a) for optimization problems involving quadratics.

We need to maximize the area function A(x) = x(20 - x) = 20x - x². This is a quadratic that opens downward (since the coefficient of x² is negative), so it has a maximum at its vertex. The x-coordinate of the vertex is x = -b/(2a) = -20/(2(-1)) = -20/(-2) = 10. We can verify this makes sense: when x = 10, the length is also 20 - 10 = 10, giving a square with maximum area. Common mistakes include choosing an endpoint (x = 0 or x = 20) instead of the vertex, or confusing perimeter with area. For optimization problems with quadratics, the extremum always occurs at the vertex unless constrained to an interval.

To write x² + 6x + 1 in vertex form, we need to complete the square. First, take half of the coefficient of x and square it: (6/2)² = 9. Add and subtract 9: x² + 6x + 9 - 9 + 1 = (x + 3)² - 8. The vertex form is (x + 3)² - 8, which can also be written as (x - (-3))² + (-8). Common errors include forgetting to subtract the added term (getting (x + 3)² + 10) or making sign errors with the h-value. When completing the square, always remember to balance what you add by subtracting the same amount.

This question asks for the equation to find x where a rectangle has area 48, length x, width x-2. Set up x(x-2)=48, so $x^2$ -2x -48=0. This is a quadratic from the area formula. Solving via factoring: (x-8)(x+6)=0, x=8 or -6 (discard negative), but the question wants the equation. Completing the square or formula isn't needed here. A key error is sign mistakes, like B's +2x giving wrong width relation. Choice D assumes a square, ignoring the x-2. Model real-world problems into quadratics and check signs carefully.

This question asks for the axis of symmetry of $f(x)=x^2$ +8x +3. For $y=ax^2$ + bx + c, the axis is x=-b/(2a), so x=-8/2=-4. To derive, complete the square: $y=(x^2$ +8x +16) -16 +3 = $(x+4)^2$ -13, vertex at x=-4. The quadratic formula isn't directly needed but roots would be symmetric around -4. A common error is forgetting the negative, choosing x=4 as in D. Choices A and B might come from misreading coefficients. Memorize the axis formula for quick identification in graphing problems.

This question asks for the equation of a parabola opening upward with vertex (-2,1) passing through (0,9). Use vertex form y = a(x + $2)^2$ + 1, and substitute (0,9): $a(2)^2$ + 1 = 4a + 1 = 9, so 4a=8, a=2, giving $y=2(x+2)^2$ +1. Since it opens upward, a>0 confirms. Completing the square isn't needed but could verify if expanded. A key error is miscalculating a, like in B where a=1 gives y(0)=5, not 9. Choices C and D shift the vertex or sign incorrectly. When given vertex and a point, plug into vertex form to solve for a quickly.

This question asks for the time (t) when the ball reaches its maximum height in the model (h(t) = $-5t^2$ + 20t + 1). For a quadratic $(at^2$ + bt + c), the vertex gives the maximum at (t = -rac{b}{2a}), so here (t = -rac{20}{2(-5)} = rac{20}{10} = 2). This is because the parabola opens downward (a = -5 < 0), and the vertex is the peak. You could complete the square: (h(t) = $-5(t^2$ - 4t) + 1 = -5(t - $2)^2$ + 21), confirming the vertex at t=2. A common error is using t = rac{b}{2a} without the negative, leading to t=-2, which isn't among choices. Choice D miscalculates as rac{20}{5}=4, ignoring the 2a denominator. On the SAT, memorize the vertex formula for projectile problems to quickly find max/min times.