### Basic form

Basic form of the quadratic equation with constant coefficients a, b and c:

$a·{x}^{2}+b·x+c=0$

$\text{with}\phantom{\rule{1em}{0ex}}a,b,c\phantom{\rule{0.5em}{0ex}}\in \phantom{\rule{0.5em}{0ex}}\mathbb{R}\phantom{\rule{1em}{0ex}}\text{und}\phantom{\rule{1em}{0ex}}a\phantom{\rule{0.5em}{0ex}}\ne \phantom{\rule{0.5em}{0ex}}0$

### Normal form

Division by the coefficients a and renaming of terms $\frac{b}{a}$ and $\frac{c}{a}$ leads to the normal form of the quadratic equation:

${x}^{2}+\frac{b}{a}x+\frac{c}{a}=0$

$\text{with}\phantom{\rule{1em}{0ex}}p=\frac{b}{a}\phantom{\rule{0.5em}{0ex}}\text{und}\phantom{\rule{0.5em}{0ex}}q=\frac{c}{a}\phantom{\rule{0.5em}{0ex}}\text{follows the basic form}$

${x}^{2}+px+q=0$

### General solution of the quadratic equation

By transforming and applying the quadratic complement, the general solution of the quadratic equation can be given in the form of the p,q formula:

Starting from the normal form of the quadratic equation, the equation is solved using quadratic supplement.

${x}^{2}+px+q=0$

Starting point for the general solution is the normal form of the quadratic equation.

${x}^{2}+px=-q$

1. Subtraction q

${x}^{2}+px+{\left(\frac{p}{2}\right)}^{2}-{\left(\frac{p}{2}\right)}^{2}=-q$

2. Expanding the equation by ${\left(\frac{p}{2}\right)}^{2}$ and subtracting this term so that the equation actually does not change.

${x}^{2}+px+{\left(\frac{p}{2}\right)}^{2}={\left(\frac{p}{2}\right)}^{2}-q$

3. After the transformation, the left-hand side of the equation contains an expression that corresponds to the 1st Binomial theorem: ${\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$.

${\left(x+\frac{p}{2}\right)}^{2}={\left(\frac{p}{2}\right)}^{2}-q$

$x+\frac{p}{2}=±\sqrt{{\left(\frac{p}{2}\right)}^{2}-q}$

5. Pull the root then allows the resolution of the equation for x. Because the square root in general has a positive and a negative quadratic equation, the solution also in general two solutions x1 und x2.

${x}_{1,2}=-\frac{p}{2}±\sqrt{{\left(\frac{p}{2}\right)}^{2}-q}$

6. Result is the so-called pq-formula for determining the solution of a quadratic equation.

The solutions can be divided into three categories depending on the value of the discriminant: $D={\left(\frac{p}{2}\right)}^{2}-q$ :

$D=0$ : There is one real solution.

$D>0$ : There are two real solutions.

$D<0$ : There are two complex solutions.

### Example of a quadratic equation with two real solutions

The first example has two real solutions. In the following, the approach is shown with a square expansion and then with the pq-formula.

${x}^{2}+3x+2=0$

Example equation

${x}^{2}+3x=-2$

Subtracting the absolute term

${x}^{2}+3x+{\left(\frac{3}{2}\right)}^{2}-{\left(\frac{3}{2}\right)}^{2}=-2$

With the addition of the term ${\left(\frac{3}{2}\right)}^{2}$ the expression is extended to the first binomial formula.

${x}^{2}+2\frac{3}{2}x+{\left(\frac{3}{2}\right)}^{2}-{\left(\frac{3}{2}\right)}^{2}=-2$

Expanding the factor in front of x illustrates the binomial structure.

${x}^{2}+2\frac{3}{2}x+{\left(\frac{3}{2}\right)}^{2}={\left(\frac{3}{2}\right)}^{2}-2$

After forming is on the left side of the equation the first binomial.

${\left(x+\frac{3}{2}\right)}^{2}={\left(\frac{3}{2}\right)}^{2}-2$

Application of the first binomial theorem ${\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$

$x+\frac{3}{2}=±\sqrt{{\left(\frac{3}{2}\right)}^{2}-2}$

Application of the square root allows the resolution of the equation for x. The square root of generally has a positive and a negative solution.

${x}_{1,2}=-\frac{3}{2}±\sqrt{{\left(\frac{3}{2}\right)}^{2}-2}$

${x}_{1}=-1$

${x}_{2}=-2$

Forming and calculating results on the two real solutions of the quadratic equation.

${x}^{2}+{3}x+{2}=0$

To the solution reaches you even by employing the coefficients of the equation in the pq-formula.

${x}_{1,2}=-\frac{{p}}{2}±\sqrt{{\left(\frac{{p}}{2}\right)}^{2}-{q}}$

p and q must be replaced by the coefficients.

${x}_{1,2}=-\frac{{3}}{2}±\sqrt{{\left(\frac{{3}}{2}\right)}^{2}-{2}}$

${x}_{1}=-1$

${x}_{2}=-2$

Use of p = 3 and q = 2 gives the solution of the equation.

### Example of a quadratic equation with two complex solutions

The second example has two complex solutions. In the following, the approach is first with square complement and then shown with the pq-formula.

${x}^{2}+1=0$

Example equation

${x}^{2}=-1$

This very simple quadratic equation can be transformed directly.

${x}_{1,2}=\sqrt{-1}=±i$

${x}_{1}=+i$

${x}_{2}=-i$

The special feature is that the discriminant is negative so the term under the square root. The square root of -1 is denoted by i. The i stands for imaginary unit.

${x}^{2}+{0}x+{1}=0$

To the solution is arrived at by substituting the coefficients of the equation in the pq-formula.

${x}_{1,2}=-\frac{{p}}{2}±\sqrt{{\left(\frac{{p}}{2}\right)}^{2}-{q}}$

p and q must be replaced by the coefficients.

${x}_{1,2}=-\frac{{0}}{2}±\sqrt{{\left(\frac{{0}}{2}\right)}^{2}-{1}}$

${x}_{1}=+i$

${x}_{2}=-i$

Inserting p = 0 and q = 1 results in the solution of the equation.

### Example of a quadratic equation with a two-fold solution

The third example has a two-fold real solution.

${x}^{2}+4x+4=0$

Example equation

${x}^{2}+4x+4-4=-4$

${\left(x+2\right)}^{2}=0$

${x}_{1,2}=-2$

The solution by completing the square leads to a discriminant with a value of 0 means that there is a twofold solution before with +/- 0.

$\left(x+2\right)\left(x+2\right)={x}^{2}+4x+4$

${x}_{1,2}=-2$

From the product representation of the equation it is apparent that it is a two-fold solution.

### The pq-formula for solving a quadratic equation

The application of pq-formula requires that the quadratic equation is in the normal form. If they do not exist so they can be converted by transformations in the normal form. Here an example the necessary transformations to normal form.

$2{x}^{2}-4x+6=2$

Example equation

${x}^{2}-2x+3=1$

Division by the factor before x2

${x}^{2}-2x+2=0$

Subtracting the right-hand side

${x}^{2}+\left(-2\right)x+2=0$

Considering the sign of p, p = -2 and q = 2 can be read off.

Calculator for the solution of the quadratic equation:

$a·{x}^{2}+b·x+c=0$

Enter the coefficients a, b and c of the quadratic equation:

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a=
b=
c=

## Vertex form

The vertex form of the square function is:

$y={\left(x-{x}_{V}\right)}^{2}+{y}_{V}$

Where xV and yV are the x and y coordinates of the vertex of the parabola. The vertex is the minimum or maximum of the function, depending on whether the parabola is up or down.

Vertex form from basic form:

In the basic form, the coefficient before x2 is 1.

Basic form of the quadratic function with the constant coefficients p and q:

$y={x}^{2}+px+q$

If the square function is in basic form, the vertex of the parabola is given by:

${x}_{V}=-\frac{p}{2}$

${y}_{V}=-{\left(\frac{p}{2}\right)}^{2}+q$

Transformation from the basic form to the vertex form with quadratic expansion and application of the first binomial:

${x}^{2}+px+q=$

${x}^{2}+px+{\left(\frac{p}{2}\right)}^{2}-{\left(\frac{p}{2}\right)}^{2}+q=$

${\left(x+\frac{p}{2}\right)}^{2}-{\left(\frac{p}{2}\right)}^{2}+q=$

${\left(x-\frac{-p}{2}\right)}^{2}-{\left(\frac{p}{2}\right)}^{2}+q$

Calculator for transform Normal form to vertex form

## Parabola

The solutions of the quadratic equation corresponding to the zeros of a parabola. A parabola is defined by a mapping of the form correspond to the zeros of the function $f\left(x\right)=a·{x}^{2}+b·x+c$ . From this follows that the solution of the quadratic equation $a·{x}^{2}+b·x+c=0$ corresponds to the zeros of the function f(x). Where the parabola intersects the x-axis are the solutions to the equation.

Depending on the location of the parabola are two zeros, one zero or no zeros. If the parabola do not intersect the x-axis has the corresponding quadratic equation complex solutions.

Interactive graphical representation of a parabola Parabola Plotter

## More Calculators

Here is a list of of further useful calculators:

Index Calculator Quadratic Equation Quadratic Equations Parabola Plotter Normal form to vertex form Derivative calculus