Proof using $\epsilon, \delta$-definition Prove that $\displaystyle \lim_{x\to-2}  (5x^2-3x+4)=30$.
I have it factored to $|5x-13||x+2| < \epsilon$.
and now I can't figure where to go from there
 A: There is a more or less standard way to approach these problems: Given $\epsilon>0$, you want to find a $\delta>0$ such that if $0<|x+2|<\delta$, then $|(5x^2-3x+4)-30|<\epsilon$. As you say, the later can be rewritten as $|5x-13||x+2|<\epsilon$. We need to find a $\delta$ that ensures this is the case.
Note that $|5x-13|=|5(x+2)-23|\le 5|x+2|+23$, using the triangle inequality. 
We now require that the $\delta$ we need to exhibit will be at most $1$. The specific number ($1$ in this case) is irrelevant. I just want a constant. Under this requirement, we see that $$5|x+2|+23<5\delta+23\le 5+23=28.$$ Hence, $$|5x-13||x+2|\le 28|x+2|<28\delta.$$ Since we want the original product to be at most $\epsilon$, the easiest way to arrange that now is to require that $28\delta$ itself is at most $\epsilon$.
In summary, we see we have imposed two requirements on $\delta$: It must be at most $1$, and it must be at most $\epsilon/28$.
We can combine both requirements and finally pick $\delta$ as $\min\{1,\epsilon/28\}$. We can now verify that this $\delta$ indeed works.
Now, note we did not pick the "optimal" value of $\delta$. In fact, we did not even attempt to find such optimal value. And there is some room for improvement. (For example, we can replace the constant $1$ with another constant, and see how that affects the value of $\delta$. To illustrate, if we only require that $\delta<2$, then all we can say now is that $5|x+2|+23<5\delta+23\le10+23=33$, so the other requirement for $\delta$ is now that it is at most $\epsilon/33$.)  
A: Let $\epsilon>0$ and take $\delta=\min(1,\epsilon)$ then if $|x-(-2)|=|x+2|\leq \delta$ then by triangle inequality we have $|x|\leq 2+\delta\leq 3$ so we find
$$|5x-13|\leq 5|x|+13\leq28$$
hence we find
$$|5x^2-3x+4-30|=|5x-13||x+2|\leq 28\delta\leq 28\epsilon$$
Added If we want the final result $\cdots\leq \epsilon$ we can rectify our proof by choosing $\delta=\min(1,\frac{\epsilon}{28})$
