Answer
The solution set is $\left\{-4, 4\right\}$.
Work Step by Step
Factor out the GCF (which is 3) on the left side to obtain:
$3(x^2-16)=0
\\3(x^2-4^2)=0$
Factor the difference of two squares using the formula $a^2-b^2=(a-b)(a+b)$
to obtain:
$3(x-4)(x+4)=0$
Use the Zero-Product Property by equating each factor with a variable to zero to obtain:
$\begin{array}{ccc}
&x-4=0 &\text{ or } &x+4=0
\\&x=4 &\text{ or } &x=-4
\end{array}$
Therefore the solution set is $\left\{-4, 4\right\}$.
