Answer
The solution set is $\left\{-3, 3\right\}$
Work Step by Step
Write 9 as $3^2$ to obtain:
$x^2-3^2=0$
Factor the difference of two squares using the formula $a^2-b^2=(a-b)(a+b)$ to obtain:
$(x-3)(x+3)=0$
Use the Zero-Product Property by equating each factor to zero and then solving the equation to obtain:
$\begin{array}{ccc}
&x-3=0 &\text{ or } &x+3=0
\\&x=3 &\text{ or } &x=-3
\end{array}$
Thus, the solution set is $\left\{-3, 3\right\}$.
