AoPS Wiki talk:Problem of the Day/September 21, 2011

We see, after substitution, that $x=\sqrt{x}+2$ and thus, isolating the square root and squaring, $x=(x-2)^2=x^2-4x+4$ and therefore $x^2-5x+4=0$ . The sum of the roots of this equation, by Vieta's formulas, is $\boxed{5}$ .