# Java Program - Find Roots of a Quadratic Equation

A standard form of a quadratic quation is:

$ax^2 + bx + c = 0$

Where:

$a, b$ and $c$ are real numbers and $a \ne 0$.

Roots of the equation are:

$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

### For Example:

The roots of equation $x^2 + 5x + 4 = 0$ is

$\frac{-5 \pm \sqrt{5^2 - 4\times1\times4}}{2\times1}$ = $\frac{-5 \pm \sqrt{9}}{2}$ = $\frac{-5 \pm 3}{2}$ = $-4, -1$

The roots of the equation will be imaginari if $D = b^2 - 4ac \lt 0$. For example - the roots of equation $x^2 + 4x + 5 = 0$ will be

$\frac{-4 \pm \sqrt{4^2 - 4\times1\times5}}{2\times1}$ = $\frac{-4 \pm \sqrt{-4}}{2}$ = $\frac{-4 \pm 2i}{2}$ = $-2 \pm i$

### Example: Calculate roots of a Quadratic equation

In the below example, a method called roots is created which takes $a$, $b$ and $c$ as arguemts to calculate the roots of the equation $ax^2 + bx + c = 0$.

import java.lang.Math;

public class MyClass {
static void roots(double a, double b, double c) {
double D = b*b - 4*a*c;
if (D >= 0){
double x1 = (-b + Math.sqrt(D))/(2*a);
double x2 = (-b - Math.sqrt(D))/(2*a);
System.out.println("Roots of the equation ax^2 + bx + c = 0 are :");
System.out.println(x1 + " and " + x2);
System.out.println("Where a, b, c are " + a + "," + b + " and " + c + " respectively.");
} else {
double x1 = -b/(2*a);
double x2 = Math.sqrt(-D)/(2*a);
System.out.println("Roots of the equation ax^2 + bx + c = 0 are imaginary.");
System.out.println("Real part of root: " + x1);
System.out.println("Imaginary part of root: " + x2);
System.out.println("Where a, b, c are " + a + "," + b + " and " + c + " respectively.");
}
}

public static void main(String[] args) {
roots(1,5,4);
System.out.println();
roots(1,4,5);
}
}


Output

Roots of the equation ax^2 + bx + c = 0 are :
-1.0 and -4.0
Where a, b, c are 1 , 5 and 4 respectively.

Roots of the equation ax^2 + bx + c = 0 are imaginary.
Real part of root: -2.0
Imaginary part of root: -1.0
Where a, b, c are 1 , 4 and 5 respectively.