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.

I assume you mean
$$ x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4\pm\ldots}}}}}$$
so that
$$ \begin{align}x&=\sqrt{4+\sqrt{4-x}}\\
 x^2&=4+\sqrt{4-x}\\
 (x^2-4)^2&=4-x\\
 0 &= x^4-8x^2+x+12= (x^2-x-3)(x^2+x-4)\end{align}$$
Since clearly $x\ge \sqrt 4=2$, the second factor is $x^2+x-4\ge2>0$, which leaves us with the positive solution $x=\frac{1+\sqrt{13}}{2}\approx 2.3027756 $ from the first factor.
