Answer
12
Work Step by Step
We are given the following equation:
$$ H=\frac{1}{2}(n^2-n)$$
Plugging in $66$ for $H$, we find:
$$ 66=\frac{1}{2}\left(n^2-n\right)\\ 132=n^2-n \\ n^2-n-132=0 \\ n_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-4\cdot \:1\left(-132\right)}}{2\cdot \:1}$$
Considering only positive values:
$$ n=12$$
