2. Given $x \in \mathbf{R}, f(x)=x^{2}+12 x+30$. Then the solution set of the equation $f(f(f(f(f(x)))))=0$ is $\qquad$ .
2. $\{-6-\sqrt[32]{6},-6+\sqrt[32]{6}\}$.

It is easy to see that $f(x)=x^{2}+12 x+30=(x+6)^{2}-6$. Then $f(f(f(f(f(x)))))=(x+6)^{32}-6$. Therefore, the equation $f(f(f(f(f(x)))))=0$ is $(x+6)^{32}-6=0 \Rightarrow x=-6 \pm \sqrt[32]{6}$.

Thus, the solution is $\{-6-\sqrt[32]{6},-6+\sqrt[32]{6}\}$.