Squaring and combining terms we get $a-x^2=\sqrt{a+x}$, squaring again we get $a^2-2ax^2+x^4=a+x$. Instead of solving this fourth degree equation in $x$, we change our view: we will solve this as a quadratic equation in $a$. Rearranging, we have $a^2-(2x^2+1)a+(x^4-x)=0$. By quadratic formula, we get $a=x^2-x$ and $a=x^2+x+1$. When $a=x^2-x\implies x^2-x-a=0$, we get $x=\frac{1+\sqrt{4a+1}}{2}$, since $x\geq 0$. When $a=x^2+x+1\implies x^2+x+1-a=0$, we get $x=\frac{-1+\sqrt{4a-3}}{2}$. Since $a-x^2=\sqrt{a+x}$ implies $x^2\leq a\implies x\leq\sqrt{a}$, it follows that $x=\boxed{\frac{-1+\sqrt{4a-3}}{2}}$ is the only solution.