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!
 A: "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.
A: As for your second question, you drop the $5$ due to the fact $n \leq 5n$ and the added $C+2$ is due to the fact $C+2 -\frac{16}{n} \leq C+2$ as $n$ is positive.
A: You have $5n^2-2n+16\leq Cn$. 
Step 1: If you divide by $n>0$ then you get $5n-2+\frac{16}n\leq C$. 

Step 2: Add $2-\frac{16}n$ to the inequality and you get $5n\leq C+2-\frac{16}n$.

Step 3: On the LHS you use $n\leq 5n$ and you get $n\leq C+2-\frac{16}n$.

Step 4: Since $n>0$ you have $-\frac{16}n<0$. Use this for the RHS of the inequality and you get $n\leq C+2$.
Your prof combined these steps to one inequality. 

Then there exist constants C and k so that 5n2−2n+16≤Cn for all n > k.

Here is the definition of $O(n)$ used and claimed that the inequality holds for large $n$. But from the inequality you concluded that $n$ has to be bounded which is a contradition.
A: The idea behind a big-O proof is to show that something you have (here $f(n)$) grows "as fast as" something simpler, typically some power of $n$.  If something is $O(n)$ then it means that it has essentially linear growth; if it's $O(n^2)$ then it has quadratic growth, etc.
Note that this growth can still appear large when you first look at it: $f(n)=10^6n+10^{78}$ will have enormous values for $n\geq 1$... but it's still linear and will be overtaken by $g(n)=n^2+6$ for $n$ large enough.
Your proof is working by contradiction: it assumes that $f(n)$ is $O(n)$ and tries to show that that leads to something obviously false.  If it's $O(n)$ then it takes the form $Cn$ for some constant $C$, so we work with that.  So we have:
\begin{eqnarray} f(n)  =  5n^2 - 2n + 16 & \leq & Cn  \quad \leftarrow \mbox{ this is our assumption} \\
 \Rightarrow 5n - 2 + 16/n & \leq & C \\
 \Rightarrow n & \leq & C+2 - 16/n
\end{eqnarray} 
because $n\leq 5n$.
As $n$ gets large $16/n \rightarrow 0$, but also $n$ can get as large as we like, while $C$ is fixed.  So there is no (finite) value of $C$ that we can choose that makes this inequation true for all $n$, so we have reached our contradiction.
When working with inequalities like this the aim isn't to balance both sides (as you do with equations), the aim is to simplify things as much as possible, and to do that you "throw things away" when you can correctly say that this will make one side smaller or larger, depending on your inequality sign.
