Answer: $1,-\frac{5}{2},-1 \pm \sqrt{3}$.

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

$$
\left\{\begin{array}{l}
4 x^{2}+7 x-10=y \\
4 y^{2}+7 y-10=x
\end{array}\right.
$$

is satisfied. Subtracting the second equation from the first, after transformations we get $4(x-y)(x+y+2)=0$, from which either $y=x$, or $y=-x-2$. Substituting $y$ into the first equation of the system. In the first case, we get the equation $4 x^{2}+6 x-10=0$, from which $x=1$ or $x=-\frac{5}{2}$. In the second case, we get the equation $4 x^{2}+8 x-8=0, x^{2}+2 x-2=0$ from which $x=-1 \pm \sqrt{3}$.

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

$\left(4\left(4 x^{2}+7 x-10\right)^{2}+7\left(4 x^{2}+7 x-10\right)-10\right)-x=64 x^{4}+224 x^{3}-96 x^{2}-512 x+320=32\left(2 x^{4}+7 x^{3}-3 x^{2}-16 x+10\right)$,

i.e., the original equation is equivalent to

$$
2 x^{4}+7 x^{3}-3 x^{2}-16 x+10=0
$$

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:

$$
2 x^{4}+7 x^{3}-3 x^{2}-16 x+10=(x-1)(2 x+5)\left(x^{2}+2 x-2\right)
$$

The roots of the third factor can be found, for example, using the quadratic formula.
Problem 5. Let $f(x)=4 x^{2}+7 x-10$. Solve the equation $f(f(x))=x$.

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