## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

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: $$n=12$$