$f(x)=\sqrt{\dfrac{x-\sqrt {20}}{x-3}}$, $g(x)=\sqrt{x^2-4x-12}$, find the domain of $f(g(x))$ 
$f(x)=\sqrt{\dfrac{x-\sqrt {20}}{x-3}}$, $g(x)=\sqrt{x^2-4x-12}$, find the domain of $f(g(x))$.

For some reason, I cannot get the correct answer. Here is what I tried.
$D_g:$
$$x^2-4x-12\ge0$$
$$(x-6)(x+2)\ge0$$
$$\boxed{(-\infty,-2] \cup [6,\infty)}$$
$D_f:$
$$\dfrac{x-\sqrt{20}}{x-3}\ge0$$
$$\boxed{(-\infty, \sqrt {20}] \cup (3,\infty)}$$
$D_{g(f)}:$
$$\sqrt{x^2-4x-12}\le \sqrt{20}$$
$$[-4,8]$$
$$\text{or}$$
$$\sqrt{x^2-4x-12}>3$$
$$(-\infty, -3) \cup (7,\infty)$$
$$\boxed{(-\infty, -3)\cup (-3,-2) \cup (7,\infty)}$$
So then my final answer was
$$\color{blue}{\boxed{(-\infty, -3)\cup (-3,-2) \cup [6,7) \cup (7,\infty)}}$$
What did I do wrong? 
Correct Answer:

$x\le-4\ \text{or}\ -3<x\le-2\ \text{or}\ 6\le x<7\ \text{or}\ x\ge8$


The domain of $f$ is $(-\infty,3)\cup[\sqrt{20},\infty)$ (you seem to believe instead that $3>\sqrt{20}$).
Thus we need
$$
\begin{cases}
x^2-4x-12\ge 0 \\[4px]
\sqrt{x^2-4x-12}<3
\end{cases}
\qquad\text{or}\qquad
\begin{cases}
x^2-4x-12\ge 0 \\[4px]
\sqrt{x^2-4x-12}\ge\sqrt{20}
\end{cases}
$$
