Big Oh where $g(n)$ is $n^2-n$
$f \in O(g)$ means there's $n_0$ and $c&gt;0$ such that $n&gt;n_0$ implies $f(n) \leq cg(n)$.

Note that for $n&gt;2$, we have $n^2 \geq 3n$.

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

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