$x^2-\sqrt{a-x}=a$ <=>$x^2-x+\frac{1}{4}=a-x+\sqrt{a-x}+\frac{1}{4}$ <=>$(x-\frac{1}{2})^2=(\sqrt{a-x}+\frac{1}{2})^2$ case 1: $x-\frac{1}{2}=\sqrt{a-x}+\frac{1}{2}$ <=>$x-1=\sqrt{a-x}$ <=>$a\ge 1$ and $x=\frac{\sqrt{4a-3}+1}{2}$ case 2: $x-\frac{1}{2}=-(\sqrt{a-x}+\frac{1}{2})$ <=>$x=-\sqrt{a-x}$ <=>$a=-1/4, x=-1/2$; $-1/4< a \leq 0, x=(+-\sqrt{4a+1}-1)/2$ or $a>0, x=(-\sqrt{4a+1}-1)/2$