Answer
$\dfrac{2}{3}$
Work Step by Step
Add $2$ to both sides:
$\sqrt {6-3x}-2+2=0+2$
$\sqrt {6-3x}=2$
Square both sides:
$(\sqrt {6-3x})^2=(2)^2$
$6-3x=4$
Subract $6$ from both sides:
$6-3x-6=4-6$
$-3x=-2$
Divide both sides by $-3$:
$\dfrac{-3x}{-3}=\dfrac{-2}{-3}$
$x=\dfrac{-2}{-3}$
$x=\dfrac{2}{3}$
Substitute into original problem
$\sqrt {6-3\left(\frac{2}{3}\right)}-2=0$
$\sqrt {6-2}-2=0$
$\sqrt {4}-2=0$
$2-2=0$
$0=0$
