## Precalculus: Mathematics for Calculus, 7th Edition

The real solutions of the equation $x^{2}+x-12$ are x=-4,3.
1. Factor $x^{2}+x-12$: The factors are $(x-3)(x+4)$ because $-3\times 4$ = -12 (the c value in $a^{2}+bx+c$ and -3+4 =1, the b value. 2. Set x-3 and x-4 equal to zero and solve: $x=3, x=-4$