JAVA利用公式x1 = (-b + sqrt(b*b-4*a*c))/(2*a), x2 = (-b - sqrt(b*b-4*a*c))/(2*a)求一元二次方程ax2 + bx + c =0的根，其中a不等于0。

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\u4ee5\u4e0b\u662f\u4ee5JAVA\u8bed\u8a00\u7f16\u5199\u7684\u4ee3\u7801\uff1a

import java.util.Scanner;

public class QuadraticEquationSolver {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.println("Enter the coefficients of the quadratic equation ax^2 + bx + c = 0:");
        double a = scanner.nextDouble();
        double b = scanner.nextDouble();
        double c = scanner.nextDouble();

        double discriminant = b * b - 4 * a * c;

        if (discriminant > 0) {
            double x1 = (-b + Math.sqrt(discriminant)) / (2 * a);
            double x2 = (-b - Math.sqrt(discriminant)) / (2 * a);
            System.out.println("The roots are real and different.");
            System.out.println("x1 = " + x1);
            System.out.println("x2 = " + x2);
        } else if (discriminant == 0) {
            double x = -b / (2 * a);
            System.out.println("The roots are real and equal.");
            System.out.println("x = " + x);
        } else {
            System.out.println("The roots are complex and different.");
            double realPart = -b / (2 * a);
            double imaginaryPart = Math.sqrt(-discriminant) / (2 * a);
            System.out.println("x1 = " + realPart + "+" + imaginaryPart + "i");
            System.out.println("x2 = " + realPart + "-" + imaginaryPart + "i");
        }
    }
}

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