Solve the following inequality I have the following inequality $\frac {2x^2}{x+2} &lt; x-2$. I tried to solve it the with the following steps.

step 1 $\frac {2x^2}{x+2} &lt; x-2$

step 2 $\frac {2x^2}{x+2} - (x-2) &lt; 0$

step 3 $\frac {2x^2}{x+2} - \frac {(x-2)(x+2)}{1(x+2)} &lt; 0$

step 4 $\frac {2x^2}{x+2} - \frac {x^2-2^2}{x+2} &lt; 0$

step 5 $\frac {2x^2 - x^2 + 4}{x+2} &lt; 0$

step 5 $\frac {x^2 + 4}{x+2} &lt; 0$

step 6 I used character study to get the result x > 2. But this is incorrect.

Where did I go wrong with this?. I feel that I made a mistake somewhere in step 2 but not sure what I did wrong.

Thanks!
HINT : $\frac{x^2+4}{x+2}\lt 0$ is correct. Then, note that $x^2+4\gt 0$ for any $x\in\mathbb R$.
Your inequality holds good if $x&gt;-2$, not $2$.