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?