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

\[P(E) = {n \choose k} p^k (1-p)^{ n-k} \]

\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}

\[\left( \sum_{k=1}^n a_k b_k \right)^{\!\!2} \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)\]