How do you solve this equation? $$\sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}= 5$$
 A: This is actually a very common problem in elementary algebra textbooks. When you do enough of them, you memorize the methods.
The important thing to note is that the braced part in the equation below is equal to the entire thing, or in other words, $5$.  
$$\sqrt{x+\underbrace{\sqrt{x+\sqrt{x+\cdots}}}}$$
This means the equation just becomes
$$\sqrt{x+5}=5$$
Can you continue from there?
A: Hint $$\sqrt{x + \color{blue}{\sqrt{x + \sqrt{x + \sqrt{x + ...}}}}}= \sqrt{x+\color{blue}{5}}$$
A: $x+n+a = \sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+(x+n)\sqrt{\ldots}}}$
This formula was proven by Ramanujan.
A: In the wiki page for nested radicals you will find the formula (for $n>0$)
$$
\sqrt{n+\sqrt{n+\sqrt{n+\ldots}}}
= \frac{1+\sqrt{1+4n}}{2}
$$
This implies
$$
\frac{1+\sqrt{1+4x}}{2}
= 5,
$$
showing $x = 20$.
This formula can be shown like this: 
Let $\phi := \sqrt{n+\sqrt{n+\sqrt{n+\ldots}}}$. Then $\phi^2 = n + \phi$. This is a quadratic in $\phi$ which one can easily solve with the quadratic formula. You have to choose the positive solution since roots of positive numbers are positive.
