When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$ Equation: \begin{equation} E = mc^2 \end{equation} Equation*: \begin{equation*} E = mc^2 \end{equation*} Brackets: \[E = mc^2\] Brackets tagged: \[E = mc^2\tag{x}\] Align: \begin{align} a_1& =b_1+c_1\\ a_2& =b_2+c_2-d_2+e_2 \end{align} Align*: \begin{align*} a_1& =b_1+c_1\\ a_2& =b_2+c_2-d_2+e_2 \end{align*} Align: \begin{align} a_{11}& =b_{11}& a_{12}& =b_{12}\\ a_{21}& =b_{21}& a_{22}& =b_{22}+c_{22} \end{align} Align with \notag and \tag: \begin{align} a_{11}& =b_{11}& a_{12}& =b_{12}\notag\\ a_{21}& =b_{21}& a_{22}& =b_{22}+c_{22} \tag{y} \end{align} Align* with \tag: \begin{align*} a_1& =b_1+c_1\tag{z}\\ a_2& =b_2+c_2-d_2+e_2 \end{align*}