For \(ax^2+bx+c=0\),
\[x={-b\pm\sqrt{b^2-4ac}}\over{2a}\]

For \(ax^3+bx^2+cx+d=0\),
\[\begin{align}
x &= \sqrt[3]{\left(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\right)-\sqrt{\left(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\right)^2+\left(\frac{c}{3a}-\frac{b^2}{9a^2}\right)^3}} \\
&+ \sqrt[3]{\left(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\right)+\sqrt{\left(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\right)^2+\left(\frac{c}{3a}-\frac{b^2}{9a^2}\right)^3}} \\
&- \frac{b}{3a}
\end{align}\]