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(Created page with "Testing MathJax: <math>\begin{align} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla...") |
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\end{align}</math> | \end{align}</math> | ||
<math>x = {-b \pm \sqrt{b^2-4ac} \over 2a}</math> | Inline <math>x = {-b \pm \sqrt{b^2-4ac} \over 2a}</math> | ||
When $a \ne 0$, there are two solutions to $ax^2 + bx + c = 0$ and they are | |||
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$ | $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$ | ||
<pdf>Rooter.pdf</pdf> | |||
Latest revision as of 13:27, 11 November 2021
Testing MathJax:
[math]\displaystyle{ \begin{align} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{align} }[/math]
Inline [math]\displaystyle{ x = {-b \pm \sqrt{b^2-4ac} \over 2a} }[/math]
When $a \ne 0$, there are two solutions to $ax^2 + bx + c = 0$ and they are
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$