Proving that $\lim_{x\to 4}\frac{2x-5}{(6-3x)(x-6)} = \frac{1}{4} $ 
Proving that $$\lim_{x\to 4}\frac{2x-5}{(6-3x)(x-6)} = \frac{1}{4} $$

I need to prove this using delta-epsilon.
$\forall \varepsilon \gt 0,\;\exists\delta \gt 0\;\forall x \in D $ (where $D$ is the domain of the function)
$$ 0 \lt |x-4|\lt \delta \Rightarrow |\frac{2x-5}{(6-3x)(x-6)} - (\frac{1}{4})| < \epsilon $$
$$ 0 \lt |x-4|\lt \delta \Rightarrow |-\frac{(3x-4)(x-4)}{12(x-6)(x-2)}| < \epsilon $$
Great so we have (x-4) in our term, that's usable since we know that |x-4|>0. 
How do I go about proving the rest of this? Do I have to leave the minus in the term or can I leave it out considering it's the absolute value?

When you have rational functions depending on $x$ where the denominator is not approaching to zero, a good idea is try to estimate the value. For instance, note that if $|x-4|<1$, then $3<x<5$ and we have 
$$\left|\dfrac{(3x-4)(x-4)}{-12(x-6)(x-2)}\right|=\left|\dfrac{(3x-4)(x-4)}{12(6-x)(x-2)}\right|\leq \dfrac{1}{12(6-5)(3-2)}\left|(3x-4)(x-4)\right|\leq \dfrac{1}{12}(3(5)-4)\left|x-4\right|=\dfrac{11}{12}|x-4|$$
If we want the last expression to be less than $\epsilon$, it will suffice to take $\delta=\frac{12}{11}\epsilon$.
Thus, you can take $\delta=\min\{1,\frac{12}{11}\epsilon\}$, and the proof will follow from the previous analysis.
