To solve an equation with an absolute value, recognize that there are two things that would make an absolute value equal 5: the inside of the absolute value could equal 5, or it could equal -5. So your job is to solve for both possibilities:

Possibility 1:

You can then subtract 3 from both sides to get:

And divide both sides by 2 to finish:

Possibility 2: 

Subtract 3 from both sides to get:

Divide both sides by 2:

Your two solutions are 1 and -4. Since the question wants the sum of the two answers, you can add them together to get your answer, -3.

When you're solving equations involving absolute values, it's important to recognize that there are generally two solutions. Here if the inside of the absolute value equals 1, you've solved for  -- or if the inside of the absolute value equals -1 you've also solved for . So you should solve this as two equations:

Possibility 1

Here you can add 7 to both sides to get: 

And then divide both sides by 2:

Possibility 2

Add 7 to both sides:

 And divide both sides by 2:

The question asks for the sum of all answers, so add  to get the right answer, 

An important thing to know about absolute values is that their minimum value is zero; absolute values must be nonnegative.  So here if you take the result of an absolute value and then add 1 to it, it simply cannot equal 0. To do so, the absolute value itself would have to equal -1, and that is just not possible.

One helpful shortcut on this problem is just understanding that the result of an absolute value can never be negative. So an answer choice like  simply cannot be correct: it's not a valid equation.

Of course, there are three remaining choices so your guessing probability isn't high enough to quit now. To solve this, first think about which values are 5 units away from -3. Since -3 + 5 = 2, and -3 - 5 = -8, you have two values that you know fit the definition: 2 and -8.

Now plug those numbers into the answer choices to see which fit with one of the absolute values. You'll see that  fits:

, so this satisfies 

, so this satisfies 

The quadratic equation  factors to . Remember, to solve for the solutions of a quadratic you can factor it to two parenthetical terms multiplied together, because then you can leverage the fact that anything times zero is zero. If either of those parentheticals were to equal zero, then the entire equation equals zero.

The most common place to make a mistake on quadratic problems is to properly factor as above but to not take the final step of setting each parenthetical term equal to zero. Here to solve you need to perform:

, so  is the solution to the first parentheses.

, so  is the solution to the second parentheses.

Any time you see  and  terms in an equation, you're likely dealing with a quadratic or polynomial, and it is therefore likely that the best way to solve those is to move all the terms to one side of the equation to set it equal to zero. Here you can do that by subtracting  from both sides to arrive at the quadratic:

Now your job is to factor the quadratic, looking for two values that multiply to  and sum to .  This should lead you to a factored quadratic of:

The final - and often missed when working under time pressure! - step is to set each parentheses equal to zero to solve the equation.  This means that your solutions are:

, so 

, so 

Since you know that  is one solution to the quadratic, and that the last term of the quadratic is , the equation must factor to:

, an equation with solutions of  and , and a quadratic that FOILS to have a  as its numerical term. If you fully FOIL out this equation, you have:

And if you then line that up with the given equation, you can see where  fits:

 must then be .

This quadratic offers a great shortcut for those fluent in the art of factoring quadratics. You know that  is one of the solutions, meaning that the quadratic must factor to:

Where you now just need to determine how to fill the second parentheses. And you know that when factoring a quadratic, you need to multiply to the last term and sum to the middle term. And here you know the middle term has a coefficient of . So in order to arrive at a sum of , the factorization must be:

, meaning that the other solution is .

Of course, there's a "long way" on this problem that isn't that much longer. If you know that  is a solution, you can plug in  to the given quadratic and solve for :

So 

Then you can plug that back into the original and factor the quadratic:

So the solutions are , which you were given, and , the answer you need.

This problem features a common quadratic type, the "Difference of Squares" structure, in which . If you quickly see that structure, you should see that  needs to be the square root of , meaning that it's  (or ).  That makes the answer .

Of course, you can also use classic quadratic factoring and FOIL to solve this, also. If you were to FOIL the algebraic expression on the right hand side of the equation, you'd get:

Note that those middle two terms cancel, leaving just . This also tells you that  needs to be the square root of , meaning that  or . Only  is listed as an answer choice, so that is your answer.

This quadratic factors to , where the numerical terms multiply to , the last term, and sum to , the middle term. But notice that the two parentheses are the same! This quadratic can be simplified even further to one of the common quadratics you should remember, . This quadratic's simplest form is:

So the only unique solution to this equation is , making  the correct answer. Note the word "unique" in the question - that should signal to you to check for duplicate solutions, like you may have arrived at when initially factoring this quadratic into two sets of parenteheses.