Big O Proof , f(n) = 2n + 1 and I have to prove f(n) is O n^2

If I have $$f(n) = 2n + 1$$ and I have to prove $$f(n) \in O(n^2)$$, by proving there exists positive constants $$c$$ and $$n_0$$ such that $$f(n), can I do this all in one step by putting in values of $$c$$ and $$n_0$$ at once or do I have to put in values of c first and then get n on one side and put in values for $$n_0$$, the for all $$n \ge n_0$$ confuses me.

It all depends on the case and how easy it is to search $$n_0$$ or use $$c$$, in your case you can show that $$2n + 1$$ belongs to $$n^2$$, taking the constant $$c = 1$$, so that in the proof of BigO, $$g (n)$$ dominates $$f(n)$$, what is the same $$2n + 1 \le c (n^2)\to 2n + 1 \le n^2$$; now you can search a $$n_0$$.
You have to demonstrate that there is one c and one $$n_0$$ such that 2n+1 <= c $$n^2$$ for all $$n >= n_0$$. In this case, 2n+1 <= $$3n^2$$ for n >= 1, 2n+1 <= 1.25 $$n^2$$ for n >= 2, 2n+1 <= 7/9 $$n^2$$ for n >= 3 etc, so there are many combinations for c, n_0.
To prove this, you usually go the easiest way possible. For example 2n+1 <= 2n^2 + n^2 = 3 n^2, so c = 3 and $$n_0$$ = 1 works. There is no need to find the smallest possible c or $$n_0$$.