College Algebra (10th Edition)

The solution set is $\left\{3\right\}$.
Square both sides to obtain: $15-2x=x^2$ Subtract $15$ and add $2x$ on both sides of the equation to obtain: $15-2x-15+2x=x^2+2x-15 \\0=x^2+2x-15$ Factor the trinomial to obtain: $(x+5)(x-3) = 0$ Equate each factor to zero, and then solve each equation to obtain: $\begin{array}{ccc} &x+5=0 &\text{ or } &x-3=0 \\&x=-5 &\text{ or } &x=3 \end{array}$ Note that $-5$ is an extraneous solution since the principal square root is never negative. Thus, the solution set is $\left\{3\right\}$.