## College Algebra (10th Edition)

The solution set is $\left\{-4, 1\right\}$.
Subtract $4$ on both sides to obtain: $x^2+3x-4=0$ We factor to obtain: $(x+4)(x-1)=0$ Use the Zero-Product Property by equating each factor to zero. Then, solve each equation to obtain: $\begin{array}{ccc} &x+4 = 0 &\text{ or } &x-1=0 \\&x=-4 &\text{ or } &x=1 \end{array}$ Thus, the solution set is $\left\{-4, 1\right\}$.