解题思路:
求根公式: x1=(-b-sqrt(b^2-4ac))/2a
x2=(-b+sqrt(b^2-4ac))/2a
注意事项:
a!=0
参考代码:
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()) {
double a = scanner.nextDouble();
double b = scanner.nextDouble();
double c = scanner.nextDouble();
double x1 = (-b-Math.sqrt(b*b-4*a*c))/2*a;
double x2 = (-b+Math.sqrt(b*b-4*a*c))/2*a;
if(x1>=x2){
System.out.printf("%.2f %.2f",x1,x2);
}
else{
System.out.printf("%.2f %.2f",x2,x1);
}
}
}
}
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