Ultimatewarrio7 wrote: Obviously, $x\geq 0,$ otherwise there aren't any real solutions. Notice : $\mathcal{LHS}$ is strictly increasing while $\mathcal{RHS}$ is strictly decreasing, thus, the given equation has at most $1$ solution, since $x=0$ is a solution it is also unique. It seems that if $x=0$ then $RHS=\sqrt{a-\sqrt{a}} \not =0$.

Seicchi28 wrote: Determine all real numbers $x$ which satisfy \[ x = \sqrt{a - \sqrt{a+x}} \]where $a > 0$ is a parameter. $\iff$ (squaring) $x\ge 0$ and $\sqrt{a+x}=a-x^2$ $\iff$ (squaring) $x\ge 0$, $x^2\le a$ and $x^4-2ax^2-x+a^2-a=0$ $\iff$ $x\in[0,\sqrt a]$ and $(x^2-x-a)(x^2+x+1-a)=0$ $x^2-x-a$ has one negative root and one positive root $>\sqrt a$, and so none fits. $x^2+x+1-a$ has a nonnegative root if and only if $a\ge 1$ (sum of roots is negative) and it is then indeed $<\sqrt a$ And so solution $\boxed{x=\frac{-1+\sqrt{4a-3}}2\text{ if and only if }a\ge 1}$