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$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$ | $x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$ | ||
[a+b] | |||
Revision as of 18:33, 1 November 2022
This page is for testing the wiki.
[math]\displaystyle{ \int_{0}^{2\pi}\!\! \int_{0}^{\infty} e^{-r^2}r\,dr\,d\theta = \pi }[/math]
$\mathbb{R}^n$
<math>\mathbb{R}^n</math>
$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$
[a+b]
$c = \sqrt{a^2+b^2}$