$f \in O(g)$ means there's $n_0$ and $c>0$ such that $n>n_0$ implies $f(n) \leq cg(n)$.

Note that for $n>2$, we have $n^2 \geq 3n$.

Then, for $n>2$, we have: 
$$
\begin{align}
2(n^2 - n) & = n^2 - 2n + n^2 \\
& \geq n^2 - 2n + 3n\\
& = n^2 + n\\
\end{align}
$$

Summarizing: for $n>2$, we have $n^2+n \leq 2(n^2-n)$. Thus $n^2+n \in O(n^2-n)$.