When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
\[ \det \hspace{3pt} \begin{array}{|cc|}t - z&x + iy\\x- iy&t + z\end{array}\hspace{3pt} = t^2 - x^2 - y^2 - z^2\]
- yes
- no
Klein Quadric
Dec 27, ’10
8:10 AM