When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
\[
\frac{\pi}{2} =
\left( \int_{0}^{\infty} \frac{\sin x}{\sqrt{x}} dx \right)^2 =
\sum_{k=0}^{\infty} \frac{(2k)!}{2^{2k}(k!)^2} \frac{1}{2k+1} =
\prod_{k=1}^{\infty} \frac{4k^2}{4k^2 - 1}
\]
\[
\frac{\pi}{2} =
\left( \int_{0}^{\infty} \frac{\sin x}{\sqrt{x}} dx \right)^2 =
\sum_{k=0}^{\infty} \frac{(2k)!}{2^{2k}(k!)^2} \frac{1}{2k+1} =
\prod_{k=1}^{\infty} \frac{4k^2}{4k^2 - 1}
\]