6.042. $x^{2}+3 x-18+4 \sqrt{x^{2}+3 x-6}=0$.
## Solution.

Let $\sqrt{x^{2}+3 x-6}=y \geq 0$. Then $x^{2}+3 x-6=y^{2}$

or

$$
x^{2}+3 x=y^{2}+6
$$

and the equation becomes:

$$
y^{2}+6-18+4 y=0 \Leftrightarrow y^{2}+4 y-12=0, y_{1}=-6 \text { (not suitable), } y_{2}=2 \text {. }
$$

Then

$$
\begin{aligned}
& \sqrt{x^{2}+3 x-6}=2 \Leftrightarrow x^{2}+3 x-6=4 \Leftrightarrow \\
& \Leftrightarrow x^{2}+3 x-10=0 ; x_{1}=-5, x_{2}=2
\end{aligned}
$$

By verification, we confirm that these are the roots of the given equation.

Answer: $x_{1}=-5, x_{2}=2$.