Proving $3n^2 - 4n \in \Omega(n^2)$ Proving $3n^2 - 4n \in \Omega(n^2)$
Attempt:
$3n^2 - 4n \geq cn^2$
$n(3n-4) \geq cn^2$
$(3n-4) \geq cn$
$3n - 4 - cn \geq 0$
$n(3-c) \geq 4$
$n \geq \frac{4}{3-c}$
would do I choose for $n_0$ and $c$ to satisfy this proof? 

$3n^2 - 4n \in \Omega(n^2)$ means:
there are $c>0$ and $n_0 \in \mathbb N$ such that $3n^2-4n \ge cn^2$ for $n \ge n_0$.
Since $\frac{3n^2-4n}{n^2}= 3-\frac{4}{n} \to 3$ as $n \to \infty$, there is $n_0$ such that $\frac{3n^2-4n}{n^2} \ge 2$ for $n \ge n_0$.
Hence $3n^2-4n \ge 2n^2$ for $n \ge n_0$.
This shows that $3n^2 - 4n \in \Omega(n^2)$.
