編程：求a㎡＋nm+c=0的根。分別考慮：①有兩個不等的實根；②有兩個相等的實根

public void equation(double a, double b, double c) {

double D = b * b - 4 * a * c; //D是判別式

if (D > 0) {

System.out.println("方程的兩個根為:");

double x1 = (Math.pow(D, 0.5) - b) / (2 * a);

System.out.println("根x1=" + x1);

double x2 = -(Math.pow(D, 0.5) + b) / (2 * a);

System.out.println("根x2=" + x2);

} else if (D < 0) {

System.out.println("方程判別式小於0");

} else {

System.out.println("方程只有壹個根為");

double x = -(Math.pow(D, 0.5) + b) / (2 * a);

System.out.println("x=" + x);

}

}