Need help proving that $f(n) = 5n^2 - 2n + 16$ is not O(n) So I have tried time and time again but do not understand how to conclude this proof. The answer to the example states:

Prove that $5n^2 - 2n + 16$ is not O(n). Assume $5n^2 - 2n + 16$ is O(n). Then there exist constants C and k so that $5n^2 - 2n + 16 \le Cn$ for all n > k. Dividing both sides by n (and assuming n > 0) we get $5n - 2 + 16/n \le C$, or $n \le C + 2 - 16/n \le C + 2$. This equality does not hold for $n > C + 2$ , contrary to our assumption that it held for all large values of n. Therefore $5n^2 - 2n + 16$ is not O(n).

Considering my professor just keeps pointing at this answer and not actually listening to my questions I will try to get clarified on here.
My first question about this proof is:
How does $n > C + 2$ disprove big O? What is the condition testing against, is it infinite in some way? and where do we ever assume that C "held for all large values of n"? and what does that even mean?
I understand this proof up until the point where

we get $5n - 2 + 16/n <= C$, or $n <= C + 2 - 16/n <= C + 2$.

in this snippet I don't understand where $5n$ disappears to and why we are left with just $n$ on the LHS? And after that, why are we left with only $C + 2$ on the RHS, where did $16/n$ go?
I may be thinking of this problem too much like an equation and trying to balance each side but I don't know any other ways to approach it. 
I am not very well versed in big-oh proofs but I am really trying to understand how they work and how to solve them for my class. If anybody could lend insight, answers to my questions, a walk through, or even hints. I just really need help understanding this proof, thanks in advance!

"How does n>C+2 disprove big O? What is the condition testing against, is it infinite in some way? and where do we ever assume that C "held for all large values of n"? and what does that even mean?" 
You started by saying "$\underline{5n^2−2n+16≤Cn}$ for all $n > k$." So in particular there is some number $k$ so that the underlined inequality  holds for all $n>k$. So as soon as $n$ is large enough, the underlined inequality holds. But you also showed that it doesn't hold when $n>C+2$. So in particular if $n$ is bigger than both $k$ and $C+2$, you get the contradiction that the underlined inequality both does and does not hold.
