Problem 5. Let $f(x)=9 x^{2}+8 x-2$. Solve the equation $f(f(x))=x$.

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Answer: $-1, \frac{2}{9}, \frac{-3 \pm \sqrt{13}}{6}$.

First solution. The number $x$ satisfies the equation $f(f(x))=x$ if and only if there exists a number $y$ such that the system

$$
\left\{\begin{array}{l}
9 x^{2}+8 x-2=y \\
9 y^{2}+8 y-2=x
\end{array}\right.
$$

is satisfied. Subtracting the second equation from the first, after transformations, we get $9(x-y)(x+y+1)=0$, from which either $y=x$ or $y=-x-1$. Substituting $y$ into the first equation of the system. In the first case, we get the equation $9 x^{2}+7 x-2=0$, from which $x=-1$ or $x=\frac{2}{9}$. In the second case, we get the equation $9 x^{2}+9 x-1=0$, from which $x=\frac{-9 \pm \sqrt{117}}{2 \cdot 9}=\frac{-3 \pm \sqrt{13}}{6}$.

Plan for the second solution. We will honestly calculate $f(f(x))-x$:

$\left(9\left(9 x^{2}+8 x-2\right)^{2}+8\left(9 x^{2}+8 x-2\right)-2\right)-x=729 x^{4}+1296 x^{3}+324 x^{2}-225 x+18=9\left(81 x^{4}+144 x^{3}+36 x^{2}-25 x+2\right)$, i.e., the original equation is equivalent to

$$
81 x^{4}+144 x^{3}+36 x^{2}-25 x+2=0
$$

Now we can find two rational roots of this equation using the Rational Root Theorem: if $p / q, (p, q)=1$ is a root of a polynomial with integer coefficients, then $p$ is a divisor of the constant term, and $q$ is a divisor of the leading coefficient. We get:

$$
81 x^{4}+144 x^{3}+36 x^{2}-25 x+2=(x+1)(9 x-2)\left(9 x^{2}+9 x-1\right)
$$

The roots of the third factor can be found using the quadratic formula.