Question

The square of a number exceeds that number by 12. What are the two possible solutions?

Answer 1

X - the number

$x^2=x+12 \\ x^2-x-12=0 \\ x^2+3x-4x-12=0 \\ x(x+3)-4(x+3)=0 \\ (x-4)(x+3)=0 \\ x-4=0 \ \lor \ x+3=0 \\ x=4 \ \lor \ x=-3$

The two possible solutions are -3 or 4.

Answer 2

-3 or 4 hope this helps