Math Calculators

ax2+bx+c=0

x =

-

6

11

±

√19i

11

or -0.54545 ± 0.39626i

There was an error with your calculation.

## Using a Quadratic Formula Calculator

This calculator is an easy-to-use tool that solves quadratic equations. In algebra, a quadratic equation is any equation that can be written in the following form:

ax²+bx+c=0

where

a≠0

To use the quadratic formula calculator, enter the values of A, B, and C into the corresponding fields and press "Calculate." The value of A cannot equal zero, while zero is an acceptable input for B and C. For real and complex roots, the calculator will utilize the quadratic formula to determine all solutions to a given equation. After using the quadratic formula, the calculator will also simplify the resulting radical to find the solutions in their simplest form.

You can solve any quadratic equation with the quadratic formula. To use the quadratic formula, you should first bring the given equation to the following form: ax²+bx+c=0. Then, the solutions can be found as follows:

$$x=\frac{-b±\sqrt{b²-4ac}}{2a}$$

The part of the equation under the square root, b²-4ac, is called the discriminant.

• If the discriminant is positive, b²-4ac>0, the equation will have two real roots.
• If the discriminant is negative, b²-4ac<0, the equation will have two complex roots since the square root of a negative number is a complex number.
• If the discriminant equals zero, b²-4ac=0, the equation will have only one root.

The quadratic equation calculator will display the solutions of the entered equations and the workflow of finding these solutions. The calculator will also calculate the discriminant and demonstrate whether it is positive, negative, or equal to zero.

## Practical examples

### Example 1 (with real roots)

2x²+3x-2=0

In this example

a=2,b=3,c=-2.

Using the quadratic formula for these values, we get:

$$x=\frac{-b±\sqrt{b²-4ac}}{2a}=\frac{-3±\sqrt{3^2-4(2)(-2)}}{2(2)}=\frac{-3±\sqrt{9--16}}{4}=\frac{-3±\sqrt{25}}{4}$$

The discriminant of this equation is positive,

b²-4ac=25>0

Therefore, the equation will have two real roots.

Now let's simplify the resulting radical:

$$x=\frac{-3±\sqrt{25}}{4}=\frac{-3±5}{4}$$

$$x=\frac{-3+5}{4}\ \ \ and\ \ \ x= \frac{-3-5}{4}$$

$$x=\frac{2}{4}\ \ \ and\ \ \ x=-\frac{8}{4}$$

$$x=\frac{1}{2}\ \ \ and\ \ \ x=-2$$

Finally

x=0.5

x=-2

### Example 2 (with complex roots)

Let's solve the following quadratic equation:

x²+2x+5=0

In this example

a=1,b=2,c=5

Using the quadratic formula for these values, we get:

$$x=\frac{-b±\sqrt{b²-4ac}}{2a}=\frac{-2±\sqrt{2^2-4(1)(5)}}{2(1)}=\frac{-2±\sqrt{4-20}}{2}=\frac{-2±\sqrt{-16}}{2}$$

The discriminant of this equation is negative,

b²-4ac=-16<0

Therefore, the equation will have two complex roots.

Now let's simplify the resulting radical:

$$x=\frac{-2±\sqrt{-16}}{2}=\frac{-2±4i}{2}=\frac{-2}{2}±\frac{4i}{2}=-1±2i$$

Finally,

x=-1+2i

x=-1-2i

### Example 3 (with one root)

Let's solve the following quadratic equation:

3x²+6x+3=0

In this example

a=3,b=6,c=3

Using the quadratic formula for these values, we get:

$$x=\frac{-b±\sqrt{b²-4ac}}{2a}=\frac{-6±\sqrt{6^2-4(3)(3)}}{2(3)}=\frac{-6±\sqrt{36-36}}{6}=\frac{-6±\sqrt0}{6}$$

The discriminant of this equation equals zero, b²-4ac=0. Therefore, the equation will have one root.

$$x=\frac{-6}{6}$$

Finally,

x=-1

## Derivation of the quadratic formula

As demonstrated above, you can use the quadratic formula to solve absolutely any quadratic equation, regardless of whether the discriminant is positive, negative, or equals zero. Now let's investigate how it can be derived. Knowing the basic principles of formula derivation can be very useful in case you forget the formula itself.

The algorithm of quadratic formula derivation is relatively straightforward and is based on the procedure of completing the square. To derive the solutions of the standard quadratic equation ax²+bx+c=0, you need to follow the steps below:

1. We have an equation:

ax²+bx+c=0

Move the constant C to the right side of the equation:

ax²+bx=-c

1. Get rid of the coefficient A next to the squared term . To do this, divide the equation by A:

$$x²+\frac{b}{a}x=-\frac{c}{a}$$

$$(\frac{b}{2a})^2$$

to both sides of the equation:

$$x²+\frac{b}{a}x+(\frac{b}{2a})^2=-\frac{c}{a}+(\frac{b}{2a})^2$$

1. The left-hand side now has the form

x²+2dx+d²

This expression can be rewritten as

(x+d)²

In our equation, d is expressed as

$$\frac{b}{2a}$$

So:

$$x²+\frac{b}{a}x+(\frac{b}{2a})^2 = \left(x+\frac{b}{2a}\right)^2$$

Substitute this into the left-hand side of our formula, and leave the right-hand side untouched for now:

$$\left(x+\frac{b}{2a}\right)^2=-\frac{c}{a}+(\frac{b}{2a})^2$$

Now the root x appears only once in the equation.

1. Extract the square root from both parts of the equation:

$$x+\frac{b}{2a}=± \sqrt{-\frac{c}{a}+(\frac{b}{2a})^2}$$

1. Move \$\frac{b}{2a}\$ to the right side of the equation:

$$x=-\frac{b}{2a}± \sqrt{-\frac{c}{a}+(\frac{b}{2a})^2}$$

1. Multiply the right side of the equation by

$$\frac{2a}{2a}$$

$$x=\frac{-b ± \sqrt{-\frac{c}{a} × (2a)^2 + (\frac{b}{2a})^2 × (2a)^2}}{2a}$$

1. Simplify the equation:

$$x=\frac{-b±\sqrt{-4ac+b²}}{2a}$$

1. As a result, we get a quadratic formula:

$$x=\frac{-b±\sqrt{b²-4ac}}{2a}$$

• The sum of the two roots of the quadratic equation is

$$\frac{-b}{a}$$

Consequently, if the discriminant of the quadratic equation b²-4ac equals zero, you can find the only root of the equation as

$$\frac{-b}{2a}$$

• The product of the two roots of the quadratic equation is

$$\frac{c}{a}$$

• The term "quadratic" comes from the Latin word "quadratus", which means "square." The equation was called quadratic since the highest power of the variable is 2, i.e., the variable is "squared."

• The quadratic formula in its present shape was described as early as 628 AD by the Indian mathematician Brahmagupta, who didn't use symbols but instead discussed the solution using words. Brahmagupta, however, described only one of the two possible solutions, omitting the important ± sign before the square root.

• The graph of a quadratic function y=ax²+bx+c is a parabola. The solutions, or roots, of the quadratic equation, are actually the coordinates of the interceptions of the graph with the x-axis. If the equation has two real roots, the graph intersects the x-axis twice. If the equation has only one root, the graph of the corresponding parabola only touches the x-axis at its maximum or minimum. If the equation has no real roots, the graph of the corresponding parabola does not intersect the x-axis at all.

• When the value of the coefficient by the squared term, A, approaches zero, the graph of the corresponding parabola becomes flatter, eventually tending to become a straight line. When a=0, the equation becomes linear, and the graphical representation of it is obviously a straight line!

• Similarly, when a>0, the parabola will be facing upwards. If a<0, the corresponding parabola will be opening downwards. If a=0, the "parabola" is flat, i.e., it is a straight line.

Quadratic equations are widely used in all areas of science. For example, in physics, quadratic equations are used to describe projectile motion.