Hello, World!

Let's do some math. When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are \[x = {-b \pm \sqrt{b^2-4ac} \over 2a}.\] Moar math!!! \[ \displaystyle r^{3} \frac{d}{d r} g{\left(r \right)} + 3 r^{2} g{\left(r \right)} \]