Answer
$x=5$
Work Step by Step
Add $\sqrt{2x+7}$:
$\sqrt{3x+2}-\sqrt{2x+7}+\sqrt{2x+7}=0+\sqrt{2x+7}$
$\sqrt{3x+2}=\sqrt{2x+7}$
Square both sides:
$(\sqrt{3x+2})^2=(\sqrt{2x+7})^2$
$3x+2=2x+7$
Subtract $2$ from both sides:
$3x+2-2=2x+7-2$
$3x=2x+5$
Subtract $2x$ from both sides:
$3x-2x=2x-2x+5$
$x=5$
Substitute into original problem to check for extraneous solutions:
$\sqrt{3(5)+2}-\sqrt{2(5)+7}=0$
$\sqrt{15+2}-\sqrt{10+7}=0$
$\sqrt{17}-\sqrt{17}=0$
$0=0$
