Let's call the number x. The square of a number would be [tex] x^{2} [/tex]. This is 12 more than the number itself. 12 more than the number is 12+x.
Putting this together we get [tex] x^{2} =x+12[/tex]
This is a quadratic equation so we set it equal to zero by subtracting x+12 from both sides. We obtain:
[tex] x^{2} -x-12=0[/tex]
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:
[tex]x-4=0[/tex] so x =4
and
[tex]x+3=0[/tex] so x=-3
The two solutions are 4 and -3
