jmc

Subtracting $\sqrt{x}$ from both sides and then squaring, we get $$x+2=(10-\sqrt{x}{)}^2=x-20\sqrt{x}+100.$$Therefore, $20\sqrt{x}=98,$ so $\sqrt{x}=\frac{98}{20}=\frac{49}{10}.$ Therefore, $x={\left(\frac{49}{10}\right)}^2=\boxed{\frac{2401}{100}},$ or $x=\mathrm{24.01.}$