#### Answer

$x=12$

#### Work Step by Step

Squaring both sides of the equation and then using the properties of equality, we obtain:
\begin{array}{l}\require{cancel}\left(
\sqrt{3x}
\right)^2=\left(
x-6
\right)^2
\\\\
3x=(x)^2+2(x)(-6)+(-6)^2
\\\\
3x=x^2-12x+36
\\\\
0=x^2+(-12x-3x)+36
\\\\
x^2-15x+36=0
\\\\
(x-12)(x-3)=0
\\\\
x=\{3,12\}
.\end{array}
Upon checking, only $
x=12
$ satisfies the original equation.