Solve the following equation for $x$,$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)$ I am not great at transposition and wolfram alpha confused me so I would like to see the steps in solving for x. 
$$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)$$
Wolfram alpha gave me two incompatible answers
$$x=\pm\sqrt{n}, \ \ \ \ \ \ \ \  \sqrt{n}\not=0$$
$$x=\pm\frac{\sqrt{n}}{\sqrt{3}}, \ \ \ \ \ \ \ \  \sqrt{n}\not=0$$
The first answer is wrong the second is correct.
Can anyone explain why the first solution is wrong and how to derive the second?

Let $z = \frac{1}{2}(n - x^2)$. Then the equation becomes $\Big(\frac{z}{x}\Big)^2 = z$ or better $z\Big(\frac{z}{x^2} - 1\Big) = 0$. Solutions are $z = 0$ and $z = x^2$. If $z = 0$, then $x = \pm \sqrt{n}$. If $z = x^2$, then $x = \pm \sqrt{n/3}$.
