jmc

We square both sides to get rid of the square root. This gives $3x+6=(x+2{)}^2=x^2+4x+4$. Moving everything to one side, we get $x^2+x-2=0=(x+2)(x-1)$. Solving, we get $x=1,-2$.

We plug both values back into the equation to see if either of them are extraneous.

For $x=1$, we get $\sqrt{3\cdot 1+6}=1+2$, which works.

For $x=-2$, we get $\sqrt{3\cdot -2+6}=-2+2$, which is also valid.

Therefore, the answers are $\boxed{-2}$ and $\boxed{1}$.