Example 4. Solve the equation

$$
3 \sqrt{x+3}-\sqrt{x-2}=7
$$
Solution. Isolate one of the roots on the left side:

$$
3 \sqrt{x+3}=\sqrt{x-2}+7
$$

By squaring both sides of the obtained equation, we have

$$
9(x+3)=x-2+14 \sqrt{x-2}+49
$$

Combining like terms and isolating the radical on the right side, we get the equation

$$
4 x-10=7 \sqrt{x-2}
$$

By squaring both sides of the obtained equation, we have

$$
16 x^{2}-80 x+100=49(x-2)
$$

i.e., the equation

$$
16 x^{2}-129 x+198=0
$$

which is a consequence of the original equation. We find the roots of this equation: $x_{1}=6, x_{2}=2 \frac{1}{16}$.

Substitute each of these roots into the original equation.

For $x=6$, we get a true numerical equality. Therefore, $x_{1}=6$ is a root of the original equation.

For $x=2 \frac{1}{16}$, the left side of the original equation equals $6 \frac{1}{2}$, and its right side equals 7. Since $6 \frac{1}{2} \neq 7$, the number $x_{2}=2 \frac{1}{16}$ does not satisfy the original equation, i.e., it is an extraneous root for it.

Thus, $x_{1}=6$ is the only root of the original equation.