Prove $\lim_{x\to2} (x^4 - 2x^3 + x + 3) = 5$ using Epsilon Delta Prove $\lim_{x\to2} (x^4 - 2x^3 + x + 3) = 5$ using Epsilon Delta
I am having difficulty finding the $\delta$ value.
Here is what I have done so far:
What I want to show:
$$\forall \epsilon > 0, \exists \delta > 0 \; such \; that \; 0<\mid x -2 \mid < \delta \implies \mid x^4 - 2x^3 + x - 2\mid < \epsilon $$
After a lot of manipulation, I was able to turn the equation to the following:
$$\mid x^4 - 2x^3 + x - 2\mid \; = \; \mid x-2 \mid \mid x-2 \mid \mid x + 1 \mid \mid x + 1 \mid + 3\mid x + 1 \mid \mid x - 2 \mid$$
I already know that $ \mid x -2 \mid < \delta$. I must find an upper bound for $\mid x + 1 \mid$.
$$\mid x + 1 \mid = \mid x + 1 -3 + 3 \mid \; \le \; \mid x - 2 \mid + 3 $$
Choose $\delta = 1$, then
$$\mid x - 2 \mid < 1 \implies \mid x -2 \mid + 3 < 4$$
Therefore $\mid x + 1 \mid < 4$
This means:
$$\mid x^4 - 2x^3 + x - 2\mid < (\delta)(\delta)(4)(4) + 3(\delta)(4) = 16\delta^2 + 12\delta = \epsilon$$
This is where I get stuck. How do I transform $16\delta^2 + 12\delta = \epsilon$ into solving for $\delta$ in terms of $\epsilon$?
I am not sure if I am over thinking the whole question or if the above is even remotely correct. Please let me know any mistakes that I have made.

Here's a very easy and brief way to do it:
Given $\epsilon>0$, we need $\delta>0$ such that if $0<|x-2|<\delta$, then $|(x^4-2x^3+x+3)-5|<\epsilon$. 
Now, 
$$|(x^4-2x^2+x+3)-5|=|(x-2)(x^3+1)|.
$$ If $|x-2|<1$, that is, $1<x<3$, then $x^3+1<3^3+1=28$, and so 
$$
|(x^4-2x^3+x+3)-5|=|x-2|(x^3+1)<28|x-2|.
$$ So if we take $\delta=\min\left\{1,\frac{\epsilon}{28}\right\}$, then 
$$
0<|x-2|<\delta\to|(x^4-2x^2+x+3)-5|=|(x-2)|(x^3+1)<\frac{\epsilon}{28}\cdot 28=\epsilon.
$$ Thus, by the definition of a limit, $\lim_{x\to 2}(x^4-2x^3+x+3)=5$.
