Calculating $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\dots}}} $ If $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\sqrt{4-\sqrt{ 4+\sqrt{4-\dots}}}}}} $
then find value of 2x-1
I tried the usual strategy of squaring and substituting the rest of series by x again but could not solve.

We have
$$-\left((x^2-4)^2-4\right)=x\iff  x^4-8x^2+x+12= (x^2+x-4)(x^2-x-3) =0$$
solve this equation and notice that $x\ge2$ we find that the acceptable answer is 
$$x=\frac12\sqrt{13}+\frac12$$
hence
$$2x-1=\sqrt{13}$$
