bla bla bla \(\mu\) bla bla bla

When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

`x = (-b +- sqrt(b^2-4ac))/(2a) .`

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Topic revision: r2 - 2020-08-30 - TWikiAdminUser
 
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