How to prove $\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=\frac {\sqrt {4a-3}-1}2$

It is CORRECT!

$x^4-2ax^2-x+a^2-a=0$ can be factorised as $(x^2-x-1)*(x^2+x+1-a)$. You'll get the desired answer from the right factor $(x^2+x+1-a)$.

It's wrong! You can try $a=0$.

Also, try $a=1$.

If your sequence converges then we need to solve the following equation $$\sqrt{a-\sqrt{a+x}}=x.$$ Let $a+x=y^2,$ where $y\geq0$.

Hence, $$y^2-x=a$$ and $$x^2+y=a,$$ where $x\geq0$ and $a\geq0.$

Thus, $$y^2-x^2-x-y=0$$ or $$(x+y)(y-x-1)=0.$$

If $x+y=0$ so $x=y=a=0$ and your formula is still wrong.

If $y=x+1$ then $x^2+x+1-a=0,$ which gives $$x=\frac{-1+\sqrt{4a-3}}{2}.$$ Now, since $x\geq0$, we have $$\frac{-1+\sqrt{4a-3}}{2}\geq0,$$ which gives $a\geq1$.

But for $a=1$ our sequence divergences and it should be $a>1$.