$x\geq 2$pt$\Leftrightarrow $6+3$\sqrt{x-2}$=2x+$\sqrt{x+6}$$\Leftrightarrow $6-2x+3$\sqrt{x-2}$-$\sqrt{x+6}$=0$\Leftrightarrow $2(3-x)+$\frac{9(x-2)-(x+6)}{3\sqrt{x-2}+\sqrt{x+6}}$=0$\Leftrightarrow $2(3-x)+$\frac{8x-24}{3\sqrt{x-2 +\sqrt{x+6}}}$=0$\Leftrightarrow $(x-3)($\frac{8}{3\sqrt{x-2}+\sqrt{x+6}}$)=0$\Leftrightarrow $$\left\{ \begin{array}{l} x-3=0\\ 2- \frac{8}{3\sqrt{x-2 +\sqrt{x+6}}}\end{array} \right.=0$$\Leftrightarrow $$\left\{ \begin{array}{l} x=3\\3\sqrt{x-2}+\sqrt{x+6}=4(*) \end{array} \right.$
$(*)$$\Leftrightarrow $$3\sqrt{x-2}$=4-$\sqrt{x+6}$ bình phương 2 lần ra nghiệm $\left\{ \begin{array}{l}x=\frac{11+3\sqrt{5}}{2}(loại do x\leq 5)\\ x=\frac{11-3\sqrt{5}}{2}(t/m) \end{array} \right.$
vậy pt có 2 nghiệm x=3 và x=$\frac{11+3\sqrt{5}}{2}$