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.

To solve this problem algebraically, first recognize that the  term means you're dealing with a quadratic. To solve for a quadratic, perform the necessary algebra to set the equation equal to 0. Here that means subtracting 24 from each side to reach:

From here, remember that your goal when factoring a quadratic is to find terms that multiply to the last term (the numerical term) and sum to the middle term (the linear  term). That should lead you to:

Then the last, very critical step, is to set each parentheses equal to zero to officially solve the problem.  That means that:

 yields a solution of 

 yields a solution of 

Of those, only  is an answer, so  is correct.

Whenever you're working with a quadratic, your goal should be to move all terms to one side of the equation so that you can set it equal to zero. Here that means adding  to each side and subtracting  from each side to arrive at:

From there you can factor, looking for terms that sum to  and that multiply to . You should arrive at:

Then you have to set each parentheses equal to zero to finish solving.  That will give you:

 or  as possible solutions. Since the question asks for the sum, add those together to get the correct answer, .

Whenever you're facing a quadratic, your best move is to move all the terms to one side of the equation so you can set it to zero. Then you can factor (or apply the quadratic formula if needed). Here that means subtracting  and  from both sides of the given equation to reach:

Now you can factor, looking for terms that multiply to  and that sum to . That should lead you to:

To finish the job, you need to set each parentheses equal to zero. That means that the solutions to this equation are:

 and 

Since the question asks for the sum of those solutions, add them together to get your answer, .

Whenever you're working with a quadratic structure, you should look to move all terms to one side of the equation to set it to zero. This here means subtracting  from each side of the equation to get:

You can then factor the equation, looking for values that multiply to  and sum to . You should arrive at:

From here, you need to set each parenthetical term to zero, meaning that your answers are:

Of those, only  appears as an answer, so that answer is correct.