Simplifying $\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt {5 +\cdots}}}}$ How to simplify the expression:
$$\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\cdots}}}}.$$
If I could at least know what kind of reference there is that would explain these type of expressions that would be very helpful.
Thank you.

Let $x=\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\dots}}}}$. 
Then $x^2=5+2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}}}}$.
So, $x^2-5=2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}}}}$
Remember that $x=\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\dots}}}}$
So, $2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}}}}$. So, $2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}}}}=2x$
This automatically makes our equation to be $x^2-5=2x$. Just solve the quadratic equation, and take the positive root. Why only positive? Think about it: how can $\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\dots}}}}$ equal a negative number? It must be positive! So that's why we only take the positive root.
$$x^2-2x-5=0$$
$$x=\dfrac{2\pm \sqrt{(-2)^2-4(1)(-5)}}{2(1)}$$
$$x=\dfrac{2\pm \sqrt{4+20}}{2}$$
$$x=\dfrac{2\pm \sqrt{24}}{2}$$
$$x=\dfrac{2\pm 2\sqrt{6}}{2}$$
$$x=1\pm \sqrt{6}$$
Since $1-\sqrt{6}$ equals a negative number, we reject that root. So, $x=1+\sqrt{6}$, which is also the answer to the original problem.
Answer:
$$\displaystyle \boxed{\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\dots}}}}=1+\sqrt{6}}$$
