How find this equation solve this equation
$$\sqrt{\sqrt{3}-\sqrt{\sqrt{3}+x}}=x$$
My try: since
$$\sqrt{3}-x^2=\sqrt{\sqrt{3}+x}$$
then
$$(x^2-\sqrt{3})^2=x+\sqrt{3}$$

Let $y = \sqrt{\sqrt3 + x}$, then the equation becomes: $\sqrt3 - y = x^2$, and $y^2 = \sqrt3 + x$.
So: 
$$\begin{align*}
y^2 - x  & = y + x^2 \\
\Rightarrow (y + x)(y - x - 1) & = 0
\end{align*}$$
And we can go from here.
