#### Answer

$x=\{ -2,1 \}$

#### Work Step by Step

Squaring both sides of the equation and using the properties of radicals, the solution to the given equation is
\begin{array}{l}\require{cancel}\left(
\sqrt{3x+6}
\right)^2=\left(
x+2
\right)^2
\\\\
3x+6=(x)^2+2(x)(2)+(2)^2
\\\\
3x+6=x^2+4x+4
\\\\
0=x^2+(4x-3x)+(4-6)
\\\\
x^2+x-2=0
\\\\
(x+2)(x-1)=0
\\\\
x=\{ -2,1 \}
.\end{array}
Upon checking, both solutions, $
x=\{ -2,1 \}
,$ satisfy the original equation.