Question

PLEEAASSEE HELP!!!

Answer 1

Let's call the number x. The square of a number would be $x^{2}$. This is 12 more than the number itself. 12 more than the number is 12+x.

Putting this together we get $x^{2} =x+12$
This is a quadratic equation so we set it equal to zero by subtracting x+12 from both sides. We obtain:

$x^{2} -x-12=0$

The left-had side factors as such:
(x-4)(x+3)=0

Since we have two expressions that when multiplied give zero either one or both equal zero. We set each equal to zero to obtain the solutions:
$x-4=0$ so x =4
and
$x+3=0$ so x=-3

The two solutions are 4 and -3