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When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
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When $a \ne $, there are two solutions to \(ax^ + bx + c = \) and they are
$$x = {-b \pm \sqrt{b^-4ac} \over 2a}.$$
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MathJax TeX Test Page
$$\left(\begin{array}{cc}a & b\\ c & c\end{array}\right)$$