# Java Program - Find Roots of a Quadratic Equation

A standard form of a quadratic equation is:

ax2 + bx + c = 0

Where:

a, b and c are real numbers and a ≠ 0.

Roots of the equation are: ### For Example:

The roots of equation x2 + 5x + 4 = 0 is The roots of the equation will be imaginary if D = b2 - 4ac < 0. For example - the roots of equation x2 + 4x + 5 = 0 will be ### Example: Calculate roots of a Quadratic equation

In the example below, a method called roots is created which takes a, b and c as arguments to calculate the roots of the equation ax2 + 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 are: "+ x1 + ", " + x2);
} else {
double x1 = -b/(2*a);
double x2 = Math.sqrt(-D)/(2*a);
System.out.println("Roots are: "+ x1 + " + " + x2 + "i");
System.out.println("Roots are: "+ x1 + " - " + x2 + "i");
}
}

public static void main(String[] args) {
System.out.println("Equation is x*x+5x+4=0");
roots(1,5,4);
System.out.println("\nEquation is x*x+4x+5=0");
roots(1,4,5);
}
}
```

The above code will give the following output:

```Equation is x*x+5x+4=0
Roots are: -1.0, -4.0

Equation is x*x+4x+5=0
Roots are: -2.0 + 1.0i
Roots are: -2.0 - 1.0i
```

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