## Precalculus (6th Edition)

The solution set is $\left\{2, 6\right\}$.
Subtract $8x-12$ to both sides of the equation to obtain: $x^2-(8x-12) = 8x-12-(8x-12) \\x^2-8x+12=0$ Factor the trinomial to obtain: $(x-6)(x-2)=0$ Use the Zero-Factor Property by equating each factor to zero, then solve each equation to obtain: \begin{array}{ccc} &x-6 &\text{or} &x-2=0 \\&x=6 &\text{or} &x=2 \end{array} Thus, the solution set is $\left\{2, 6\right\}$.