Example 26. Solve the equation

$$
\sqrt{x+5}+\sqrt{2 x+8}=7
$$
Solution. Solving the system of inequalities

$$
\left\{\begin{array}{r}
x+5 \geqslant 0 \\
2 x+8 \geqslant 0
\end{array}\right.
$$

we find the domain of the equation (12): $x \geqslant-4$. We have:

$$
\text { (12) } \begin{aligned}
\Leftrightarrow & \left\{\begin{array}{l}
x \geqslant-4 \\
x+5+2 \sqrt{x+5} \sqrt{2 x+8}+2 x+8=49
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x \geqslant-4, \\
2 \sqrt{(x+5)(2 x+8)}=3(12-x)
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x \geqslant - 4 , } \\
{ x \leqslant 1 2 , } \\
{ 4 ( x + 5 ) ( 2 x + 8 ) = 9 ( 1 2 - x ) ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
-4 \leqslant x \leqslant 12, \\
x^{2}-288 x+1136=0
\end{array}\right.\right.
\end{aligned}
$$

Solving the equation $x^{2}-288 x+1136=0$, we find its roots: $x_{1}=284$ and $x_{2}=4$. The condition $-4 \leqslant x \leqslant 12$ is satisfied only by $x_{2}=4$. Therefore, $x=4$ is the only root of equation (12).