Long Division

When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$ There may be one (repeated) real root, two (distinct) real roots, or complex roots (complex conjugates), which are \(a \pm b i \), where \(i = \sqrt{-1} \).

For \(a, b, c \in \mathbb{C}\):