finding the real values of $x$ such that : $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$ How to find the real values of $x$ such that : $$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$

Elaborating some on what @Fredrik Meyer suggested, one can get:
$$\begin{align*}x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}\hspace{5pt}&\Rightarrow\hspace{5pt} x^2=2+\sqrt{2-\sqrt{2+x}}\hspace{5pt}\Rightarrow\hspace{5pt} x^4-4x^2+4=2-\sqrt{2+x}\\
&\Rightarrow\hspace{5pt}x^8+16x^4+4-8x^6+4x^4-16x^2=2+x\\
&\Rightarrow\hspace{5pt} x^8-8x^6+20x^4-16x^2-x+2=0
\end{align*}$$
The last polynomial can be factored by noticing that it vanishes at $x=2$ and at $x=-1$ into $(x-2) (x+1) (x^3-3 x+1) (x^3+x^2-2 x-1)=0$ (using WA in the end).
Now you can find many roots - but beware: not all of them solve the initial problem: each transition above gives additional assumptions on $x$: first, $x\geq 0$. Then $x^2-2\geq0$ and after that $x^4-4x^2+2\leq0$. Do you see why?
All of those together imply that $\sqrt2\leq x\leq\sqrt{2+\sqrt2}$.
Using some real analysis (or WA :)), one can show that there exists a unique root of that polynomial on $(\sqrt2,\sqrt{2+\sqrt2})$.
NB: This is not the cleanest approach and not very elegant, but it works. I'm almost sure that I have seen much nicer way to solve it, but I can't think of one now.
