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1. Overview

In this article, we’ll see how to compute the solutions of a quadratic equation in Java. We’ll start by defining what a quadratic equation is, and then we’ll compute its solutions whether we work in the real or the complex number system.

2. The Solutions of a Quadratic Equation

Given real numbers a ≠ 0, b and c, let’s consider the following quadratic equation: ax² + bx + c = 0.

2.1. The Roots of a Polynomial

The solutions of this equation are also called the roots of the polynomial ax² + bx + c. Thus, let’s define a Polynom class. We’ll throw an IllegalArgumentException if the a coefficient is equal to 0:

public class Polynom {

    private double a;
    private double b;
    private double c;

    public Polynom(double a, double b, double c) {
        if (a==0) {
            throw new IllegalArgumentException("a can not be equal to 0");
        this.a = a;
        this.b = b;
        this.c = c;

    // getters and setters

We’ll solve this equation in the real number system: for this, we’ll look for some Double solutions.

2.2. Complex Number System

We’ll also show how to solve this equation in the complex number system. There is no default representation of a complex number in Java, so we’ll create our own. Let’s give it a static method ofReal to easily convert real numbers. This will be helpful in the following steps:

public class Complex {

    private double realPart;
    private double imaginaryPart;

    public Complex(double realPart, double imaginaryPart) {
        this.realPart = realPart;
        this.imaginaryPart = imaginaryPart;

    public static Complex ofReal(double realPart) {
        return new Complex(realPart, 0);

    // getters and setters

3. Calculate the Discriminant

The quantity Δ = b² – 4ac is called the discriminant of the quadratic equation. To calculate b squared in java, we have two solutions:

  • multiply b by itself
  • use Math.pow to raise it to the power of 2

Let’s stick with the first method and add a getDiscriminant method to the Polynom class:

public double getDiscriminant() {
    return b*b - 4*a*c;

4. Get the Solutions

Depending on the value of the discriminant, we’re able to know how many solutions exist and compute them.

4.1. With a Strictly Positive Discriminant

If the discriminant is strictly positive, the equation has two real solutions, (-b – √Δ) / 2a and (-b + √Δ) / 2a:

Double solution1 = (-polynom.getB() - Math.sqrt(polynom.getDiscriminant())) / (2 * polynom.getA());
Double solution2 = (-polynom.getB() + Math.sqrt(polynom.getDiscriminant())) / (2 * polynom.getA());

If we work in the complex number system, we then just need to make the conversion:

Complex solution1 = Complex.ofReal((-polynom.getB() - Math.sqrt(polynom.getDiscriminant())) / (2 * polynom.getA()));
Complex solution2 = Complex.ofReal((-polynom.getB() + Math.sqrt(polynom.getDiscriminant())) / (2 * polynom.getA()));

4.2. With a Discriminant Equal to Zero

If the discriminant is equal to zero, the equation has a unique real solution -b / 2a:

Double solution = (double) -polynom.getB() / (2 * polynom.getA());

Similarly, if we work in a complex number system, we’ll transform the solution in the following way:

Complex solution = Complex.ofReal(-polynom.getB() / (2 * polynom.getA()));

4.3. With a Strictly Negative Discriminant

If the discriminant is strictly negative, the equation has no solution in the real number system. However, it can be solved in the complex number system: the solutions are (-b – i√-Δ) / 2a and its conjugate (-b + i√-Δ) / 2a:

Complex solution1 = new Complex(-polynom.getB() / (2* polynom.getA()), -Math.sqrt(-polynom.getDiscriminant()) / 2* polynom.getA());
Complex solution2 = new Complex(-polynom.getB() / (2* polynom.getA()), Math.sqrt(-polynom.getDiscriminant()) / 2* polynom.getA());

4.4. Gather the Results

To sum up, let’s build a method that will fill in a List with the solutions of the equation when they exist. In the real number system, this method looks like this:

public static List<Double> getPolynomRoots(Polynom polynom) {
    List<Double> roots = new ArrayList<>();
    double discriminant = polynom.getDiscriminant();
    if (discriminant > 0) {
        roots.add((-polynom.getB() - Math.sqrt(discriminant)) / (2 * polynom.getA()));
        roots.add((-polynom.getB() + Math.sqrt(discriminant)) / (2 * polynom.getA()));
    } else if (discriminant == 0) {
        roots.add(-polynom.getB() / (2 * polynom.getA()));
    return roots;

If we work in a complex number system, we’ll rather write:

public static List<Complex> getPolynomRoots(Polynom polynom) {
    List<Complex> roots = new ArrayList<>();
    double discriminant = polynom.getDiscriminant();
    if (discriminant > 0) {
        roots.add(Complex.ofReal((-polynom.getB() - Math.sqrt(discriminant)) / (2 * polynom.getA())));
        roots.add(Complex.ofReal((-polynom.getB() + Math.sqrt(discriminant)) / (2 * polynom.getA())));
    } else if (discriminant == 0) {
        roots.add(Complex.ofReal(-polynom.getB() / (2 * polynom.getA())));
    } else {
        roots.add(new Complex(-polynom.getB() / (2* polynom.getA()), -Math.sqrt(-discriminant) / 2* polynom.getA()));
        roots.add(new Complex(-polynom.getB() / (2* polynom.getA()), Math.sqrt(-discriminant) / 2* polynom.getA()));
    return roots;

5. Conclusion

In this tutorial, we’ve seen how to solve a quadratic equation in Java, whether we work with real or complex numbers.

As always, the code can be found over on GitHub.

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