Nested square-root evaluating

Let's replace f(x) with y. By squaring the whole thing, we get $y^2=x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{...}}}}$ . But $\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{...}}}}$ is the same as y. This means that $x+y=y^2$ . We can replace x for 20 and get the solution of 5.

\[\begin{align} f(20) &= \color{Blue}{\sqrt{20 + \sqrt{20 + \sqrt{20 + \sqrt{\ldots}}}}} \\ f(20) &= \sqrt{20+ \color{Blue}{f(20)}} \\ f(20)^2 &= 20 + f(20) \\ f(20) &= -4, 5 \quad \quad \quad [f(20) \ne -ve] \\ f(20) &= \boxed{20}

\end{align}\]