To answer questions like your conjecture, use the definition of $O$. Paraphrasing from wikipedia at time of writing:
$f \in O(g)$ iff $f(x) \leq M\times g(x) $ for all $ x > x_0 $, for some constants $M$ and $x_0$.
So if $f(n) \in O(2n^2)$, then $f(n) \leq C \times 2n^2 \ \ \forall n > n_0$ for some constants $C$ and $n_0$ (I'm calling it $C$ instead of $M$ to avoid confusion later). You'll notice we can combine the $2C$ into one constant, which tells you $O(2n^2) = O(n^2)$.
What can we say about $f(n)/2$? We can now use normal algebra on the inequality, such as dividing both sides by 2 to get:
$f(n)/2 \leq Cn^2 \ \ \forall n > n_0$
Is this still $O(n^2)$? To check this, we need to find constants $M$ and $x_0$ such that the definition above holds. In this case, $M = C$ and $x_0 = n_0$ works. Notice that $M = 327.6C$ is also valid; you can use whatever method you want to find these constants, as long as you know they are constant.
In the real world you'll be able to "shortcut" most of this when doing your analysis, as NP-hard does in his answer, but those are just shortcuts for writing it out rigorously like this (which is what I would expect on homework specifically about $O$). If you're not sure whether it's valid, get the $O$s out by using the definition, perform algebra on the inequality, then get it back into $O$ by finding the appropriate constants (or simply showing that they do exist).