## Precalculus: Mathematics for Calculus, 7th Edition

The real solutions of the equation $x^{2}-7x+12=0$ are $x=3$ and $x=4$.
1. Factor $x^{2}-7x+12$: The factors are (x−3)(x-4) because −3×-4 = 12 (the c value in $a^{2}+bx+c$) and -3+-4 =-7, the b value. 2. Set the factors equal to zero: $x-3=0$ and $x-4=0$ 3. Solve: $x=3$ and $x=4$