[tex]Domain:\\\\\left\begin{array}{ccc}3x\geq0\to x\geq0\\x+6\geq0\to x\geq-6\end{array}\right\}\text{therefore}\ x\geq0\to x\in[0,\ \infty)\\\\------------------------\\\\\sqrt{3x}=\sqrt{x+6}\qquad\text{square of both sides}\\\\(\sqrt{3x})^2=(\sqrt{x+6})^2\qquad\text{use}\ (\sqrt{a})^2=a\ \text{for}\ a\geq0\\\\3x=x+6\qquad\text{subtract x from both sides}\\\\2x=6\qquad\text{divide both sides by 2}\\\\\boxed{x=3}\in D[/tex]
