Answer
$ \omega=6-\sqrt{52}, \omega = 6+\sqrt{52}.$
Work Step by Step
Use the completing the square method. $$
\begin{aligned}
\omega^2-12 \omega-16 & =0 \\
\omega^2-12 \omega+6^2 & =16 \mp 6^2 \\
(\omega-6)^2 & =52 \\
\omega-6 & = \pm \sqrt{52} \\
\omega & =6 \pm \sqrt{52}.
\end{aligned}
$$ This gives: $$
\begin{aligned}
\omega & =6-\sqrt{52} \\
& \approx -1.21 \\
\omega & =6+\sqrt{52} \\
& \approx13.21.
\end{aligned}
$$ The solutions are: $$\begin{aligned}
\omega&=6-\sqrt{52}\approx -1.21\\
\omega& =6+\sqrt{52}\approx 13.21.
\end{aligned}$$ Define the following version of the function and plot it. We see that the zeros are where they should be.
$$
f(\omega) =(\omega-6)^2-52.
$$
