Quadratic formula 

<math>ax^2+bx+c=0</math>

<math>

The term b² – 4ac is called the discriminant of the quadratic equation, because it discriminates between three qualitatively different cases:

• If the discriminant is zero, then there is a repeated solution x, and this solution is real. (Geometrically, this means that the parabola described by the quadratic equation touches the x-axis in a single point.)
• If the discriminant is positive, then there are two different solutions x, both of which are real. (Geometrically, this means that the parabola intersects the x-axis in two points.)
• If the discriminant is negative, then there are two different solutions x, both of which are complex numbers. The two solutions are complex conjugates of each other. (In this case, the parabola doesn't intersect the x-axis at all.)

Note that when computing roots numerically, the usual form of the quadratic formula isn't ideal. See Loss of significance for details.

Derivation

<math>ax^2+bx+c=0</math>

Dividing our quadratic equation by a, we have

<math>

which is equivalent to

<math>x^2+\frac{b}{a}x=-\frac{c}{a}.</math>

The equation is now in a form in which we can conveniently complete the square[?]. To "complete the square" is to add a constant (i.e., in this case, a quantity that doesn't depend on x) to the expression to the left of "=", that will make it a perfect square trinomial of the form x² + 2xy + y². Since "2xy" in this case is (b/a)x, we must have y = b/(2a), so we add the square of b/(2a) to both sides, getting

<math>x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}.</math>

The left side is now a perfect square; it is the square of (x + b/(2a)). The right side can be written as a single fraction; the common denominator is 4a². We get

<math>\left(x+\frac{b}{2a}\right)^2=\frac{-4ac+b^2}{4a^2}=\frac{b^2-4ac}{4a^2}.</math>

Taking square roots of both sides yields

<math>x+\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a}.</math>

Subtracting b/(2a) from both sides, we get

<math>x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.</math>

Generalizations

The formula and its proof remain correct if the coefficients a, b and c are complex numbers, or more generally members of any field whose characteristic isn't 2. (In a field of characteristic 2, the element 2a is zero and it is impossible to divide by it.)

The symbol

<math>\pm \sqrt {b^2-4ac}</math>