Solve the following equation:

$$
\sqrt{2 x+7}=\frac{x^{2}-7}{2} .
$$
The left side is meaningful exactly when $x \geq-\frac{7}{2}$; the right side is non-negative when $|x| \geq \sqrt{7}$. These two conditions are satisfied simultaneously when

$$
-\frac{7}{2} \leq x \leq-\sqrt{7} \quad \text { or } \quad \sqrt{7} \leq x
$$

Assuming (1), we square the equation, and after rearranging, we get

$$
x^{4}-14 x^{2}-8 x+21=0
$$

It is easy to verify that -3 and 1 are solutions to this, so we can factor out the $(x+3)$ and $(x-1)$ root factors:

$$
(x+3)(x-1)\left(x^{2}-2 x-7\right)=0
$$

This is clearly satisfied if $x+3=0, x-1=0$, or $x^{2}-2 x-7=0$. Thus, the solutions to (2) are:

$$
x_{1}=-3, \quad x_{2}=1, \quad x_{3}=1+2 \sqrt{2}, \quad x_{4}=1-2 \sqrt{2}
$$

Of these, only $x_{1}$ and $x_{3}$ satisfy condition (1), so the solutions to the original equation are -3 and $1+2 \sqrt{2}$.