$\sqrt{53+20\sqrt{7}}$ can be written in the form $a+b\sqrt{c}$, where $a,$ $b,$ and $c$ are integers and $c$ has no factors which is a perfect square of any positive integer other than 1. Find $a+b+c$.
We make $a+\sqrt{d}=\sqrt{53+20\sqrt{7}}$. Squaring both sides, we get: \begin{align*}
a^2+2a\sqrt{d}+d=(a^2+d)+\sqrt{4a^2 \cdot d}=53+20\sqrt{7}=53+\sqrt{2800}\\
\end{align*}We set the terms with radicals equal to each other, and ones without radicals equal.  From this, we get that $a^2+d=53$ and $\sqrt{4a^2 \cdot d}=\sqrt{2800}$, so $4a^2 \cdot d =2800$. Solving, we get that $a=5$, and $d=28$.

Therefore, $\sqrt{53+20\sqrt{7}}=5+\sqrt{28}=5+2\sqrt{7}$. $a=5$, $b=2$, and $c=7$. $a+b+c=5+2+7=\boxed{14}$.