Solution to $\sqrt{\sqrt{x + 5} + 5} = x$ There are natural numbers $a$, $b$, and $c$ such that the solution to the equation
\begin{equation*}
\sqrt{\sqrt{x + 5} + 5} = x
\end{equation*}
is $\displaystyle{\frac{a + \sqrt{b}}{c}}$. Evaluate $a + b + c$.
I am not sure where I saw this problem. My guess is that it was from a high school math competition. The solution to the equation is $\frac{1 + \sqrt{21}}{2}$. This suggests use of the quadratic formula. 
The solution set to the given equation is a subset of the solution set to
\begin{equation*}
x^{2} - 5 = \sqrt{x + 5} ,
\end{equation*}
\begin{equation*}
x^{4} - 10x^{2} + 25 = x + 5
\end{equation*}
\begin{equation*}
x^{4} - 10x^{2} - x + 20 = 0 .
\end{equation*}
Using the quartic equation (or Wolfram), the solutions to this equation are computed to be
\begin{equation*}
\frac{1 \pm \sqrt{21}}{2} , \qquad  \frac{-1 \pm \sqrt{17}}{2} .
\end{equation*}

Note that if $x = \sqrt{x+5}$ then $x = \sqrt{\sqrt{x+5}+5}$.
So, try solving $x = \sqrt{x+5}$. This is a quadratic.
