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?

The inequation is equivalent to
$$x-\frac1{x+2}=\frac{x^2+2x-1}{x+2}=\frac{(x-(-1-\sqrt2))(x-(-1+\sqrt2))}{x-(-2)}<0.$$
For the expression to be negative, you need an odd number of negative factors.
$$\begin{array}{l|ccccccc}
&&-1-\sqrt2&&-2&&-1+\sqrt2\\
\hline x-(-1-\sqrt2)&\color{green}-&0&+&+&\color{green}+&+&+\\
x-(-2)&\color{green}-&-&-&0&\color{green}+&+&+\\
x-(-1+\sqrt2)&\color{green}-&-&-&-&\color{green}-&0&+\\
\hline
&\color{green}-&0&+&|&\color{green}-&0&+\\
\end{array}$$
