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An algebraic equation of the second degree. The general form of a quadratic equation is $$\tag{1} ax^2+bx+c=0,\quad a\ne0.$$ In the field of complex numbers a quadratic equation has two solutions, expressed by radicals in the coefficients of the equation: $$\tag{2} x_{1,2} = \frac{-b \pm\sqrt{b^2-4ac}}{2a}.$$ 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 $$x^2+px+q=0$$ formula (2) has the form $$x_{1,2}=-\frac{p}{2}\pm\sqrt{\frac{p^2}{4}-q}.$$ The roots and coefficients of a quadratic equation are related by (cf. Viète theorem): $$x_1+x_2=-\frac{b}{a},\quad x_1x_2=\frac{c}{a}.$$ 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 (2) holds also if the coefficients belong to a field with characteristic different from $2$.

Formula (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).

#### References

 [a1] K. Rektorys (ed.) , Applicable mathematics , Iliffe (1969) pp. Sect. 1.20

Over a field of characteristic 2 (cf. Characteristic of a field), the solution by completing the square is no longer available. Instead, by a change of variable, the equation may be written either as $$X^2 + c = 0$$ or in Artin--Schreier form $$X^2 + X + c = 0 \ .$$
In the first case, the equation has a double root $c^{1/2}$. In the Artin--Schreier case, the map $A:X \mapsto X^2+X$ is two-to-one, since $A(X+1) = A(X)$. If $\alpha$ is a root of the equation, so is $\alpha+1$. See Artin-Schreier theorem.