What are the roots of this equation? (Quadratic Equation) $\frac{x}{\sqrt{1-x^2}}=A \Rightarrow \frac{x^2}{1-x^2}=A^2$
$\Rightarrow x^2=A^2(1-x^2)=A^2-A^2x^2$
$\Rightarrow x^2+A^2x^2=A^2$
$\Rightarrow x^2(1+A^2)=A^2$
$\Rightarrow x^2(1+A^2)-A^2=0$
I have tried this;
$\Delta=b^2-4ac=-4(1+A^2)(-A^2)=4(1+A^2)(A^2)$
$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{\pm2A\sqrt{1+A^2}}{{2(1+A^2)}}$
$x_{1,2}=\frac{\pm A\sqrt{1+A^2}}{1+A^2}$
But I got nothing.

Your algebra is fine.  Another way of writing your answer (in a more simplified form) would be
$$x=\pm\frac{A}{\sqrt{1+A^2}}$$
since $$\frac{\sqrt{t}}t=\frac1{\sqrt{t}}.$$
