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An algebraic equation of the second degree. The general form of a quadratic equation is \begin{equation}\label{eq:1} ax^2+bx+c=0,\quad a\ne0. \end{equation} In the field of complex numbers a quadratic equation has two solutions, expressed by radicals in the coefficients of the equation: \begin{equation}\label{eq:2} x_{1,2} = \frac{-b \pm\sqrt{b^2-4ac}}{2a}. \end{equation} When $b^2>4ac$ both solutions are real and distinct, when $b^2<4ac$, they are complex (complex-conjugate) numbers, when $b^2=4ac$ the equation has the double root $x_1=x_2=-b/2a$.
For the reduced quadratic equation \begin{equation} x^2+px+q=0 \end{equation} formula \eqref{eq:2} has the form \begin{equation} x_{1,2}=-\frac{p}{2}\pm\sqrt{\frac{p^2}{4}-q}. \end{equation} The roots and coefficients of a quadratic equation are related by (cf. Viète theorem): \begin{equation} x_1+x_2=-\frac{b}{a},\quad x_1x_2=\frac{c}{a}. \end{equation} The expression $b^2-4ac$ is called the discriminant of the equation. It is easily proved that $b^2-4ac=(x_1-x_2)^2$, in accordance with the fact mentioned above that the equation has a double root if and only if $b^2=4ac$. Formula \eqref{eq:2} holds also if the coefficients belong to a field with characteristic different from $2$.
Formula \eqref{eq:2} follows from writing the left-hand side of the equation as $a(x+b/2a)^2+(c-b^2/4a)$ (splitting of the square).