I have to prove that:

$$f(n)=100n+5 \neq \Omega(n^2)$$
What I tried: let's assume that $f(n)=100(n)+5= \Omega(n^2)$. Thus, there must exist some positive constant $c$ and $n\_0$ such that, 
 $$0 \leq c(n^2) \leq f(n)$$
 $$0 \leq c(n^2) \leq 100n+5$$
 $$c(n^2)-100n-5 \leq 0$$
 Now the above equation holds only if $c \lt 0$, which contradicts our previous assumption. Therefore we can say that
$$f(n)=100n+5 \neq \Omega(n^2)$$

Is this approach wrong? If it is, then what could be a better approach?