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kinerity
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test

Post by kinerity » Fri Dec 01, 2017 12:42 pm

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

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Re: test

Post by kinerity » Fri Dec 01, 2017 1:39 pm

\({5 \choose 2} = \frac{5 !}{2 ! \cdot 3 !}\)

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