So I'm trying to prove that $f = \Omega(g)$ with the following:
$f(n) = \sqrt{7n^2 + 2n - 8}$, so I simplify this out and get: $f(n) = n\sqrt{7} \sqrt{1+\frac{2}{7n}-\frac{8}{7n^2}}$. So going from there I can see that my $c$ = $\frac{a_k}{2}=\frac{\sqrt{7}}{2}$, and since I have negative coefficients my $n_0=\frac{2|A|}{\sqrt{7}}=\frac{2\sqrt{8}}{\sqrt{7}}=4\sqrt{\frac{2}{7}}$.
I have $g(n) = n\sqrt{7}$ in this case.
But I get stuck here because I know that $\sqrt{1+\frac{2}{7n}-\frac{8}{7n^2}} \ge 0$. But then to go further along with the inequality I can't multiply both sides by $n\sqrt{7}$ since one side is still $0$. So I don't get how to prove it from there? What am I missing, I've read through my textbook multiple times at this point.
In the definition "$f = \Omega(g)$ iff there are $C, n_0$ such that $f(n) \ge C g(n)$ for all $n \ge n_0$" you get to choose $n_0$. From looking at where the square root is defined, you've determined that $n_0$ should be at least $4 \sqrt{\frac27}$, but if that's not working out, you should feel free to choose a bigger $n_0$.
I want to emphasize that there's never any requirement to find the best $C$ or the best $n_0$ to use. The $\Omega$ asymptotic notation specifically exists to handwave that away.
In this case, if we take $n_0 = 10$ (for example) then we can say that for $n \ge n_0$, $$ 7 + \frac2n - \frac8{n^2} \ge 7 - \frac8{n^2} \ge 7 - \frac8{10^2} = 6.92. $$ By taking $C = \sqrt{6.92}$, this is the start of a proof that $f = \Omega(n)$. Alternatively, by taking $C = \sqrt{\frac{6.92}{7}}$, this is the start of a proof that $f = \Omega(n \sqrt7)$. (Since we are free to choose the constant $C$ however we like, we usually don't talk about $\Omega(n \sqrt 7)$, since it's not any different from $\Omega(n)$.)
You should also be aware that an alternative definition of $f = \Omega(g)$ is via limits: $f = \Omega(g)$ iff $\lim_{n \to \infty} \frac{f(n)}{g(n)} > 0$ (including the case where the limit is $+\infty$). This might be easier to think about if you're comfortable with limits, though when you do homework for a specific class you might be stuck using the definition the class uses.