Solve the following equation:

$$
x^{2}+5 x+4=5 \sqrt{x^{2}+5 x+28}
$$
$$
x^{2}+5 x+4=5 \sqrt{x^{2}+5 x+28}
$$

If we square both sides of (1), a term with $x^{4}$ will appear on the left side, resulting in a fourth-degree equation. Therefore, squaring does not lead to the goal, let's try something else. Notice that the expression under the square root and the left side of (1) differ by 24, and the difference does not involve the unknown. It is therefore useful to consider the expression under the square root as the unknown. Rewriting (1) accordingly,

$$
\left(x^{2}+5 x+28\right)-5 \sqrt{x^{2}+5 x+28}-24=0
$$

This is a quadratic equation in $\sqrt{x^{2}+5 x+28}$. Since by convention the square root is always non-negative, from this we get that

$$
\sqrt{x^{2}+5 x+28}=8
$$

which holds precisely when $x^{2}+5 x+28=64$, that is, when $x=4$ or $x=-9$. Thus, the solutions to the equation are: $x=4$ and $x=-9$.