$\text{Let }y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} $, what is the nearest value of $y^2 - y$?

Let's let

$$y=\sqrt{5+\sqrt{5-{\sqrt{5+{\sqrt5-\cdots}}}}}$$

and

$$z=\sqrt{5-\sqrt{5+{\sqrt{5-{\sqrt5+\cdots}}}}}$$

and assume that both limits exist. Note that

$$y^2=5+z\quad\text{and}\quad z^2=5-y$$

Hence

$$\begin{align} y^2-y&=(5+z)-y\\ &=5-(y-z)\\ &=5-{y^2-z^2\over y+z}\\ &=5-{(5+z)-(5-y)\over y+z}\\ &=5-{z+y\over y+z}\\ &=4 \end{align}$$

Added 5/6/14: I thought I'd go ahead and add a proof that the two limits actually do exist. I'm not sure this is the simplest proof -- I'll be happy to see something slicker -- but here goes.

Define the sequence of $y_n$'s by $y_n=\sqrt{5+\sqrt{5-y_{n-1}}}$ with $y_1=\sqrt{5+\sqrt{5}}$, and let $z_n=\sqrt{5-y_n}$. It's easy to show (by induction) that $2\lt y_n\lt4$ for all $n$, so it suffices to show that the sequence of $y_n$'s has a limit.

I'll do so by showing that it's a Cauchy sequence, and I'll do that by showing that $\sum|y_n-y_{n-1}|$ converges.

We have $y_n^2-5=\sqrt{5-y_{n-1}}$, hence

$$y_n^2-y_{n-1}^2=\sqrt{5-y_{n-1}}-\sqrt{5-y_{n-2}}={y_{n-2}-y_{n-1}\over\sqrt{5-y_{n-1}}+\sqrt{5-y_{n-2}}}$$

hence

$$|y_n-y_{n-1}|={|y_{n-2}-y_{n-1}|\over|y_n+y_{n-1}|\left(\sqrt{5-y_{n-1}}+\sqrt{5-y_{n-2}}\right)}\lt{|y_{n-1}-y_{n-2}|\over|2+2|(1+1)}={1\over8}|y_{n-1}-y_{n-2}|$$

where we've used the inequalities $2\lt y_n\lt 4$ (for all $n$) in the denominator. It follows that $\sum|y_n-y_{n-1}|$ is bounded by a geometric series with ratio $1/8$, hence converges.

Added 6/29/16: For a nice example of what can go wrong if you assume a limit exists but don't bother to check if it actually does, see this problem.

You were on the right track. Put

$$x=\sqrt{5+\sqrt{5-\sqrt{5+\ldots}}}\implies x^2=5+\sqrt{5-\sqrt{5+\ldots}}\implies$$

$$x^4-10x^2+25=5-\sqrt{5+\sqrt{5-\ldots}}\implies$$

$$x^4-10x^2+20=-\sqrt{5+\sqrt{5-\ldots}}=-x\implies\;\ldots\ldots$$