While solving this equation, $ a + b^2 = \pi $ I stumbled upon ...
When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
$\displaystyle 2-\dfrac{2}{1-\left(\dfrac{2}{\smash{2-\frac{2}{x^2}}}\right)}$
When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
$\displaystyle 2-\dfrac{2}{1-\left(\dfrac{2}{\smash{2-\frac{2}{x^2}}}\right)}$
