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.

So $x_{1,2}=\frac{\pm A\sqrt{1+A^2}}{1+A^2}$.
That's it.  Go home.  Eat lunch. You are done.
Well, not entirely.  $\sqrt{1-x^2}$ must not equal $0$ and $1-x^2\ge 0$.  Need to check that this holds for $x_{1,2}$.
$\sqrt{1-x^2} = 0 \iff x^2 = 1\iff x=\pm 1$.  And $1-x^2 \ge 0 \iff x^2 \le 1 \iff |x| \le 1$.  So we must show $|x_{1,2}| < 1$.
$|A| = \sqrt {A^2} < \sqrt {1+A^2}$. So $|A|\sqrt{1+A^2} < \sqrt{1+A^2}^2 =1+A^2$
So $|x_{1,2}| = \frac {|A|\sqrt{1+A^2}}{1+A^2} < \frac {1+A^2}{1+A^2} = 1$.
Also you can simplify $x_{1,2} = \frac {\pm A}{\sqrt{1+A^2}}$ but that is not required.
