#### Answer

The solution set is $\left\{3\right\}$.

#### Work Step by Step

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 $\left\{3\right\}$.