Finding the derivative using quotient rule... $$\frac{\text{d}}{\text{dt}}\dfrac{2(t+2)^2}{(t-2)^2}$$
I applied the quotient rule:
$$\dfrac{[2(t+2)^2]'(t-2)^2-2(t+2)^2[(t-2)^2]'}{(t-2)^4}$$
$$\dfrac{4(t+2)(t-2)^2-2(t+2)^22(t-2)}{(t-2)^4}$$
$$\dfrac{4(t+2)(t-2)-4(t+2)^2}{(t-2)^3}$$
This was part of a problem where I needed to find the second derivative of a parametric curve but I am stuck on finding this derivative. I typed this problem into wolfram alpha and it gave me $\dfrac{-(16 (x+2))}{(x-2)^3}$, I've been working on this problem for the past hour and can't figure out what I am doing wrong, could someone please explain how to do this?

Write $\displaystyle\frac{t+2}{t-2}$ as $$ 1 + \frac{4}{t-2}~,$$  and use the chain rule.  
Thus 
$$\begin{align*} \frac{\text{d}}{\text{dt}}\dfrac{2(t+2)^2}{(t-2)^2}  & = 2\frac{\text{d}}{\text{dt}}\left(1 + \frac{4}{t-2} \right)^2 \\
& = 4\left(1 + \frac{4}{t-2}\right) \left(\frac{-4}{(t-2)^2}\right)\\
& = -\frac{16(t+2)}{(t-2)^3}~.
\end{align*}$$
