3. (7 points) Let $f(x)=x^{2}+12 x+30$. Solve the equation $f(f(f(f(f(x)))))=0$.

#
# Solution.

Rewrite the expression for $f(x)$ as $(x+6)^{2}-6$.

Then

$$
\begin{gathered}
f(f(x))=(\underbrace{(x+6)^{2}-6}_{f(x)}+6)^{2}-6=(x+6)^{4}-6 \\
f(f(f(x)))=(\underbrace{(x+6)^{4}-6}_{f(f(x))}+6)^{2}-6=(x+6)^{8}-6
\end{gathered}
$$

Thus, we have:

$$
f(f(f(f(f(x)))))=(x+6)^{32}-6
$$

Therefore, the original equation is equivalent to the equation $(x+6)^{32}-6=0$, from which $x=-6 \pm \sqrt[32]{6}$.

Answer. $x=-6 \pm \sqrt[32]{6}$.