Answer
$x=2$
Work Step by Step
$\sqrt{x+2}=x$
$(\sqrt{x+2})^{2}=x^{2}$
$x+2=x^{2}$
$0=x^{2}-x-2$
$x^{2}-x-2=0$
$(x+1)(x-2)=0$
$x+1=0$ or $x-2=0$
$x=-1$ or $x=2$
We need to confirm the solutions.
First, we confirm $x=-1$:
Left side=$\sqrt{-1+2}=\sqrt{1}=1$
Right side=$-1$
The sides disagree, so $x=-1$ is not a solution.
Next, we confirm $x=2$:
Left side=$\sqrt{2+2}=\sqrt{4}=2$
Right side=$2$
The sides agree, so $x=2$ is a solution.