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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
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
![][matrix]
[matrix]: http://latex.codecogs.com/svg.latex?\begin{bmatrix}1&x&x2\1&y&y2\1&z&z^2\\end{bmatrix}
