一元二次方程可能有两个实数解，或者一个实数解，或者无实数解。

根据输入的三个系数 a、b、c 的判断解的情况：

#include <stdio.h>
#include <math.h>

int main(void)
{
    double a ,b , c;
    double delat;
    double x1, x2;
    char ch;
    do
    {

        
        
        printf("请输入一元二次方程的三个系数：\n");
        
        printf("请输入系数 a = \t");
        scanf("%lf",&a);
        
        printf("请输入系数 b = \t");
        scanf("%lf",&b);
        
        printf("请输入系数 c = \t");
        scanf("%lf",&c);
        
        delat = b*b-4*a*c;
        
        if(delat>0)
        {
            x1= ( -b+sqrt(delat) )/2;
            x2= ( -b-sqrt(delat) )/2;
            printf("有2个实数解：x1 = %lf  x2 = %lf\n",x1,x2);
        }
        else if(delat==0)
        {
            x1=( -b+sqrt(delat) )/2;
            printf("有2个相等实数解：x1=x2 = %lf\n",x1);
        }
        else
            printf("无实数解\n");

        printf("是否继续吗：Y/N\n");
        scanf(" %c",&ch);
    }while(ch=='y'||ch=='Y');
    return 0;
}