2. The inequality is equivalent to $\displaystyle \frac{x^2-4x-4}{x^2+4x-4}\geq 0$
$\displaystyle \begin{array}{c | c c c c c c c c c c c} x & -\infty & -2-2\sqrt{2} & 2-2\sqrt{2} & -2+2\sqrt{2} & 2+2\sqrt{2} & \infty\\\hline\ x^2-4x-4 & + & + & +0- & - & -0+ & +\\\hline x^2-4x-4 & + & +0- & - & -0+ & + & +\\\hline E(x) & + & +|- & -0+ & +|- & -0+ & +\\ \end{array}$
Then $\displaystyle x\in(-\infty,-2-2\sqrt{2})\cup[2-2\sqrt{2},-2+2\sqrt{2})\cup[2+2\sqrt{2},\infty)$