Question

If seven times a number is added to twice its square the result is 4. Find the numbers.

Answer 1

We'll use x to represent "the number", so therefore the equation will be:
7x + 2x^2 = 4
2x^2 + 7x = 4, subtract 4 from both sides
2x^2 + 7x - 4 = 4 - 4
2x^2 + 7x - 4 = 0
Factor it, so to break it down multiply the 2 from 2x^2 with the -4, to get x^2 and -8, so
x*2 + 7x - 8
Factoring that, you should get (x + 8) (x - 1), now you are going to want to divide it by the 2 from the original equation of 2x^2 + 7x - 4, so it'll look like this: (x + 8 / 2) (x - 1 / 2), to get (x + 4) (2x - 1). Now lastly set them equal to zero, such as:
x + 4 = 0 ; 2x - 1 = 0, you will get your solution.

I hope this helps, I know it's a lot of work.