Check my workings: Prove the limit $\lim\limits_{x\to -2} (3x^2+4x-2)=2 $ using the $\epsilon,\delta$ definition. Prove the limit $\lim\limits_{x\to -2} (3x^2+4x-2)=2 $ using the $\epsilon,\delta$ definition.

Precalculations
My goal is to show that for all $\epsilon >0$, there exist a $\delta > 0$, such that
  $$0<|x+2|<\delta\ \ \text{implies}\  |3x^2+4x-2-2|<\epsilon$$
$|3x^2+4x-2-2|=|3(x+2)^2-8x-16|$
$=|3(x+2)^2-4(x+2)|$
$\leq3|x+2|^2+4|x+2|$ by triangle inequality
$<3\delta^2+4\delta$
Hence, it is sufficient to show that $3\delta^2+4\delta=\epsilon$

Proof
For all $\epsilon>0$, choose $\delta=\min\left(\sqrt{\dfrac{\epsilon}{6}},\dfrac{\epsilon}{8}\right)$
$$\begin{align*}0<|x+2|<\delta\ \ \to\ \ &|3x^2+4x-2-2|<3\delta^2+4\delta\\&<3\left(\sqrt{\frac{\epsilon}{6}}\right)^2+4\delta\\&=\frac{\epsilon}{2}+4\delta\\&<\frac{\epsilon}{2}+4\frac{\epsilon}{8}\\&=\frac{\epsilon}{2}+\frac{\epsilon}{2}\\&=\epsilon\end{align*}$$


Therefore proven? Hehe. Not sure this will work or not. 
My doubts lies in the steps. 

Hence, it is sufficient to show that $3\delta^2+4\delta=\epsilon$
choose $\delta=\min\left(\sqrt{\dfrac{\epsilon}{6}},\dfrac{\epsilon}{8}\right)$

And hey, I am looking out for other possible ways to do this question too.
 A: You made a mistake here:
$$|3x^2+4x-2-2|=|3(x+2)^2-8x-16|=|3(x+2)^2-4(x+2)|$$
It should be $\,8\,$ instead $\,4\,$ in the RHS. All the rest you did is fine, fixing this little mistake. 
I show you now how'd I do it:
$$|3x^2+4x-2-2|=|3(x+2)^2-8(x+2)|=$$
$$|x+2|\,|3x-2|\stackrel{\text{for}\,|x+2|<0.5\Longrightarrow |3x-2|<10}<10|x+2|$$
Thus, we're fine if
$$10|x+2|<\epsilon\Longrightarrow |x+2|<\frac{\epsilon}{10}$$
Thus we can choose 
$$\delta =\min\left(\frac{\epsilon}{10}\,,\,\frac{1}{24}\right)$$
A: Here's how I would do it:
For any $\epsilon>0$, choose $\delta=\min\left(1,\dfrac{\epsilon}{11}\right)$. Then:
$$\begin{align*}
0<|x+2|<\delta\ \ \implies\ \ |(3x^2+4x-2)-2| &= |3x^2+4x-4| \\
&= |3x-2||x+2| \\
&= 3\left|x-\frac{2}{3}\right||x+2| \\
&= 3\left|x+2-\frac{8}{3}\right||x+2| \\
&\le 3\left(|x+2|+\left|\frac{-8}{3}\right|\right)|x+2| \text{ }\text{ }\text{ by triangle  ineq.}\\
&< 3\left(1 + \frac{8}{3}\right)|x+2| \text{ }\text{ }\text{ since }|x+2|<\delta\le1\\
&=11|x+2| \\
&<11\left(\frac{\epsilon}{11}\right) \text{ }\text{ }\text{ since }|x+2|<\delta\le\frac{\epsilon}{11}\\
&=\epsilon\end{align*}$$
