Java求一元二次方程的根

求一元二次方程的根。

【问题描述】编写程序，从键盘输入三个系数ax²+bx+c=0，计算方程的解并输出。需要考虑方程有虚根、方程有实根、方程是一元一次方程、没有根等。输出方程的解并保留6位小数。

import java.text.DecimalFormat;
import java.util.*;
public class  Equation{
      public static void main(String[] args) {
            Scanner reader = new Scanner(System.in);
            System.out.println("请输入方程的系数a、b、c：");
            int a, b, c;
            a = reader.nextInt();
            b = reader.nextInt();
            c = reader.nextInt();
            DecimalFormat df = new DecimalFormat("0.000000");
            int delta;
            double x1, x2;
            delta = b * b - 4 * a * c;
            if (a == 0 && b == 0 && c == 0) {
                System.out.println("方程有无限根。");
                return;
                }
                if (a == 0 && b == 0 & c != 0) {
                    System.out.println("方程无解。");
                    return;
                    }
                    if (delta == 0) {
                        x1 = b / -2.0 / a;
                        x2 = b / -2.0 / a;
                        System.out.println("方程有两个相同实根, x1 =  "
                        + df.format(x1) + ",  x2 = " + df.format(x2));
                        }
                     if (delta > 0) {
                          x1 = (-b + Math.sqrt(delta)) / 2.0 / a;
                          x2 = (-b - Math.sqrt(delta)) / 2.0 / a;
                          System.out.println("方程有两个不同实根, x1 = "
                          + df.format(x1) + ",  x2 = " + df.format(x2));
                          }
                     if (delta < 0) {
                          double s, x;
                          s = -b / 2.0 / a;
                          x = Math.sqrt(-delta) / 2.0 / a;
                          System.out.println("方程有虚根, x1 = "
                          + df.format(s) + " + " + df.format(x) + "i,  x2 = "
                          + df.format(s) + " - " + df.format(x) + "i");
                          }
            }
 }

运行结果：