How to solve $x<\frac{1}{x+2}$ Need some help with:
$$x<\frac{1}{x+2}$$
This is what I have done:
$$Domain: x\neq-2$$
$$x(x+2)<1$$
$$x^2+2x-1<0$$
$$x_{1,2} = \frac{-2\pm\sqrt{4+4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{-2\pm2\sqrt{2}}{2}$$
What about now?

Multiply the inequality by $\;(x+2)^2>0\;$ ( obviously, $\;x\neq-2\;$) :
$$x(x+2)^2<x+2\iff x^3+4x^2+3x-2<0\iff$$
$$\iff (x+2)(x+1-\sqrt2)(x+1+\sqrt2)<0\iff \color{red}{x<-1-\sqrt 2}\;\;\text{or}\;\color{red}{-2<x<-1+\sqrt2}$$
