6.064. $\sqrt{x+2}+\sqrt{3 x+8}=\sqrt{2 x+6}$.

6.064. $\sqrt{x+2}+\sqrt{3 x+8}=\sqrt{2 x+6}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}x+2 \geq 0, \\ 3 x+8 \geq 0, \\ 2 x+6 \geq 0\end{array} \Leftrightarrow x \geq-2\right.$.

Write the equation in the form $\sqrt{x+2}-\sqrt{2 x+6}=-\sqrt{3 x+8}$ and square both sides:

$$
\begin{aligned}
& x+2-2 \sqrt{(x+2)(2 x+6)}+2 x+6=3 x+8 \Leftrightarrow \\
& \Leftrightarrow \sqrt{(x+2)(2 x+6)}=0
\end{aligned}
$$

from which $x+2=0, x_{1}=-2$, or $2 x+6=0, x_{2}=-3$ - does not satisfy the domain of definition. By checking, we confirm that $x=-2$ is a root of the given equation.

Answer: $x=-2$.