Quadratic Equation Solver :   ax^2+bx+c=0
Coefficients may be real or complex numbers.

## How to use this calculator ?

This calculator is a quadratic equation solver (ax2+bx+c=0).
It calculates the exact solutions when they exist and also gives a numerical approximation of them.

### "Include complex solutions" field

Choose,
• No: if you search only real solutions.
• Yes: if you want to extend the search to complex numbers.

### Coefficients input

Here are some hints to help you enter the coefficients of the equation.
1. Accepted inputs are,
• integers, example: 5, -7
• fractions, example: 1/3 or -2/9
• decimal values, example: 3.9 or -9.65
• constants, example: pi or e
• common functions, for example: sin(pi/5)
• square root operator, example : input sqrt(3) or 3^0.5 for sqrt(3)
• complex numbers, example : 1+i ou -i
2. To enter a product of two factors, use the * operator. For example: enter 2*pi and not 2pi.

## How to solve a quadratic equation?

We suppose the equation coefficients are real numbers and we search real solutions.
We also assume that, in all of the following, a != 0.

• Step 1: Calculate the equation discriminant \Delta = b^2-4*a*c
• Step 2: Calculate the solutions

• If \Delta < 0: there is no solution.

• If \Delta = 0: there is only one solution

x_1 = -b/(2*a)

• If \Delta > 0: there are 2 solutions

x_1 = (-b-sqrt(\Delta))/(2*a)

x_2 = (-b+sqrt(\Delta))/(2*a)

## Tips and Tricks

It is useful to know some tips and tricks to quickly solve a second degree equation.
• If b is an even integer, calculations can be simplified by calculating a reduced discriminant.

• \Delta' = b'^2-a*c with b' = b/2

If this discriminant is positive, then the solutions are,

x_1 = (-b'-sqrt(\Delta'))/a

x_2 = (-b+sqrt(\Delta'))/a

• If a and c have opposite signs (e.g. a=3 and c=-5) then the equation has two real solutions because in this case, a*c < 0 and then -4*a*c > 0 which implies \Delta = b^2-4*a*c > 0

• It may be useful to quickly calculate s = a+b+c the coefficients sum because, if s=0, then x=1 is a solution of the equation.

Indeed, by replacing x with 1, we get

a*1^2+b*1+c=a+b+c=0.

The second solution can be found by factoring the equation ( x-1 is obviously a factor) or by using the following property.

• It may be useful to know the formulas of the sum and the product of the solutions of a quadratic equation.

We denote x1 and x2 the roots of the equation. S and P are the sum and product of these solutions, that is,

S = x_1+x_2

P = x_1*x_2

Then we have,

S = -b/a

P = c/a

Application: to calculate two numbers given their sum S and their product P then simply solve the following quadratic equation,

x^2 - S*x +P=0

If they exist, the two searched numbers are the roots of this equation.

## Case of Non-real solutions

If \Delta < 0, we saw that the equation has no real solution. It has in fact two complex solutions which are:

x_1 = (-b-i*sqrt(-\Delta))/(2*a)

x_2 = (-b+i*sqrt(-\Delta))/(2*a)