$x-2\sqrt{x+2}=10$ `(x≥-2,x≥10)`
$⇔x+2-2\sqrt{x+2}+1-1-2-10=0$
$⇔(\sqrt{x+2}-1)^2-13=0$
$⇔(\sqrt{x+2}-1-\sqrt{13})(\sqrt{x+2}-1+\sqrt{13})=0$
$⇔\sqrt{x+2}-1-\sqrt{13}=0$ hoặc $\sqrt{x+2}-1+\sqrt{13}=0$
$⇔x+2=(1+\sqrt{13})^2$ hoặc $x+2=(1-\sqrt{13})^2$
$⇔x+2=1+2\sqrt{13}+13$ hoặc $x+2=1-2\sqrt{13}+13$
$⇔x=12+2\sqrt{13}$(thỏa) hoặc $x=12-2\sqrt{13}$(không thỏa)
Vậy `S={12+2\sqrt{13}}`
