Find the value of $x$ such that $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$ Find the value of $x$, $$\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$$
Help guys please, I have tried and I got, $x=-2, x=1$, and I think it's wrong

This should be a comment but then too long ...
Picking up on the solution given by Jan Erlang: 
$$\left(x^4-8 x^2+10\right)^2 \left(x^4-8 x^2+14\right)^2+x-4=\left(x^2-x-3\right) \left(x^2+x-4\right) \left(x^{12}-24
   x^{10}-x^9+228 x^8+16 x^7-1087 x^6-88 x^5+2720 x^4+191 x^3-3380
   x^2-136 x+1633\right)$$
Now $x^2-x-3=0$ yields: $\frac{1}{2} \left(1\pm\sqrt{13}\right)$ and $x^2+x-4=0$ yields $\frac{1}{2} \left(-1\pm\sqrt{17}\right)$ (where only positive roots are acceptable). Further $$x^{12}-24
   x^{10}-x^9+228 x^8+16 x^7-1087 x^6-88 x^5+2720 x^4+191 x^3-3380
   x^2-136 x+1633=0$$ yields $6$ more positive roots.
