Solving for $x$ when $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x$ Suppose $x$ is a real number such that $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x.$ Find $x.$

I first squared to get rid of the first square root, which gave me $\sqrt{3} - \sqrt{\sqrt{3} + x} = x^2.$ However, I'm not sure how to move on from there. Can someone give me a hint?

Comment:
If you put $x=\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} $ under the third radical you get:
$$\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + \sqrt{\sqrt{3} - \sqrt{\sqrt{3} + \sqrt{\sqrt{3} - \sqrt{\sqrt{3} + }} . . .}} }} = x$$
Suppose there is no negative sign then by squaring both sides you get:
$$X^2=\sqrt 3 +X$$
Which gives:
$$X=\frac{-1 ±\sqrt{1+4\sqrt 3}}{2}≈0.9≈\frac{\sqrt3}{2}$$
So $x<X≈\frac{\sqrt 3}2$
