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do you have a picture?

Im assuming its something like this:            (I cant be sure I dont want to solve the wrong question.)
 

Question: What if the store brought in 100% above its previous sales in other months? Then, what would the answer be?

Factor the following:
2 a^2 - 5 a - 12

Factor the quadratic 2 a^2 - 5 a - 12. The coefficient of a^2 is 2 and the constant term is -12. The product of 2 and -12 is -24. The factors of -24 which sum to -5 are 3 and -8. So 2 a^2 - 5 a - 12 = 2 a^2 - 8 a + 3 a - 12 = a (2 a + 3) - 4 (2 a + 3):
a (2 a + 3) - 4 (2 a + 3)

Factor 2 a + 3 from a (2 a + 3) - 4 (2 a + 3):
(2 a + 3) (a - 4)

Thanks guest.  Also thanks to EP and Hectictar for the earlier answers.

I will put time into digesting this when I get a long moment.

Thanks!   

As follows:

No Problem dreamdriven! Good Luck with future math problems!

I appreciate everyone's input. Thanks.

Thank you so much guest, I didn't see that mistake! I thought dreamdriven rounded because s/he is dividing by 6.66666666666666666666667. I wasn't thinking straight- I was multitasking. Once again, Thank You!

Yep!

OMG THANK YOU GUEST! I fully understand the problem now! :)

 ok, so i figured that you didn't need to solve for x... whoops... and you could use the dicriminant...

I know, so is there a better way to solve it? How would I solve it?

You need to use more explanation, I have no idea what you are talking about.

Hi Tom,

You do not need to appologise for posting 2 questions, it is reasonably clear that you are using this forum as an an aid to your learning.

Off hand I do not see why there would need to be any quadratics with integer roots. Especially when we do not even know how many kids are in the class.

I could easily be wrong and I do not know how to show it one way or the other.

Since it could be a class of hundreds of students it is impossible to write out all the possibilities.

what i did...

Huh??

It doesn't actually matter what the solution to this quadratic is. The only thing that is important is that we find all values  such that this quadratic (in ) has a real solution. For that we look at the discriminant.

We see that we can only find a suitable value of x if \( y^2 + 4y \ge 0\)

It is on the x-axis , so the y coordinate will be '0'

x, 0

Distance from point a=    sqrt ((-2-x)^2 + 0^2)

 equate to the distance from point b

         sqrt(0-x)^2 + (4-0)^2

x^2 +16 =  4 +4x + x^2

12 = 4x

x = 3

since y = 0    the point is     3,0    and the distance from each point is 5