Show that 6n^2 + 12n is O(n^2)
Let $f(n)=6n^2 + 12n$

The $O$ notation for $f(n)$ can be derived from the following simplification rules:


If $f(n)$ is a sum of several terms, we keep only the one with largest growth rate.
If $f(n)$ is a product of several factors, any constant is omitted.


From rule 1, $f(n)$ is a sum of two terms, the one with largest growth rate is the one with the largest exponent as a function of $n$, that is: $6n^2$

From rule 2, $6$ is a constant in $6n^2$ because it does not depend on $n$, so it is omitted.

Then: $f(n)$ is $O(n^2)$