关于一元二次方程的棘手问题，高手进

以下是引用付政委在2011-4-5 23:00:12的发言：

import java.util.*;

class JieHanShu {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        double a, b, c, x1, x2, d;
        a = scan.nextDouble();
        b = scan.nextDouble();
        c = scan.nextDouble();
        if (a == 0)
        {
            if (b == 0)
            {
                if (c == 0)
                    System.out.println("方程的根是任意实数");
                else
                    System.out.println("方程无根");
            }
            else
            {
                System.out.println("x=" + (-b / c));
            }
        }
        else
        {
            d = b * b - 4 * a * c;
            if (d > 0) {
                System.out.println("方程有两个实根");
                x1 = (-b + Math.sqrt(d)) / 2 / a;
                x2 = (-b - Math.sqrt(d)) / 2 / a;
                System.out.println("x1=" + x1);
                System.out.println("x2=" + x2);
            } else if (d == 0) {
                System.out.println("方程有一个实根");
                x1 = -b / 2 / a;
                System.out.println("x1=" + x1);
            } else {
                System.out.println("方程有两个虚根");
                System.out.println("x1=" + (-b / 2 / a) + "+" + Math.sqrt(-d)
                        / 2 / a + "i");
                System.out.println("x1=" + (-b / 2 / a) + "-" + Math.sqrt(-d)
                        / 2 / a + "i");
            }
        }
    }
}