Using the $\varepsilon$-$\delta$ definition of a limit, show that $\lim_{x\to 2}\frac{x^3+2x^2-8x}{x-2}=12$

Using the $\varepsilon$-$\delta$ definition of a limit, show that$$\lim_{x\to2}\frac{x^3+2x^2-8x}{x-2}=12$$

My first thought it to remove the singularity, by factorising out $(x-2)$ and getting$$\lim_{x\to2}x(x+4)=12$$$$|x(x+4) - 12|=0$$From there, I would evaluate $δ$ for each value of $x$, and have an answer of the form $δ = \min(ε,ε)$, but I am unsure of how to proceed.