Answer
$x=4$
Work Step by Step
$\sqrt x+2=x$
$\sqrt x=x-2$
$(\sqrt x)^2=(x-2)^2$
$x=x^2-4x+4$
$0=x^2-5x+4$
$x^2-4x-x+4=0$
$x(x-4)-(x-4)=0$
$(x-1)(x-4)=0$
$x-1=0$ or $x-4=0$
$x=1$ or $x=4$
Check for extraneous solutions:
$\sqrt {1}+2=3,1=1$ (eliminate this solution)
$\sqrt 4+2=4$ (keep this solution)
