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When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

Will result in:

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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3. mdx-math