3. [4 points] Solve the equation $\left(\frac{x}{\sqrt{2}}+3 \sqrt{2}\right) \sqrt{x^{3}-4 x+80}=x^{2}+10 x+24$.
Answer: $4, \sqrt{13}-1$.

Solution. Factoring the right-hand side and multiplying both sides by $\sqrt{2}$, we get $(x+6) \sqrt{x^{3}-4 x+80}=\sqrt{2}(x+4)(x+6)$. From here, there are two possibilities: either $x+6=0$ (then $x=-6$, which does not fit the domain of definition, as the expression under the square root is negative), or $\sqrt{x^{3}-4 x+80}=\sqrt{2}(x+4)$. We solve the latter equation:

$$
\left\{\begin{array}{l}
x^{3}-4 x+80=2(x+4)^{2}, \\
x+4 \geqslant 0
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{3}-2 x^{2}-20 x+48=0, \\
x \geqslant-4
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x-4)\left(x^{2}+2 x-12\right)=0 \\
x \geqslant-4
\end{array}\right.\right.\right.
$$

The equation in the system has roots $x=4, x=-1 \pm \sqrt{13}$, and of these, the inequality is satisfied by $x=4$ and $x=-1+\sqrt{13}$. These are the answers to the problem.