#include<math.h>
void main()
{
    float a=0.0,float b=0.0,float c=0.0, float disc;
    double x1=0.000000,x2=0.000000;
    printf("input 3 numbers:");
    scanf("%f%f%f",&a,&b,&c);
    disc=b*b-4*a*c;
    if(a==0.0)
    if(b==0.0)
    printf("no answer due to input error\n");
    else printf("the single root is %f\n",-c/b);
    else
    {if(disc>=0.0)
     {
     if(disc>0.0)
     x1=(-b+sqrt(disc))/(2*a);
     x2=(-b-sqrt(disc))/(2*a);
     printf("%f%f",x1,x2);
     else
     {x1=x2=-b/(2*a);
     printf("%f%f",x1,x2);}
     }
     else
     printf("无解");
   
}  
不知道怎样改错，寻求帮助

#include<stdio.h>
#include<math.h>
void main()
{float a,b,c,disc;
 double x1,x2;
    printf("input 3 numbers:");
    scanf("%f,%f,%f",&a,&b,&c);
    disc=b*b-4*a*c;
    if(a==0.0)
      printf("the single root is %f\n",-c/b);
    else
    if(disc<0) printf("无解");
    else
       {   x1=(-b+sqrt(disc))/(2*a);
           x2=(-b-sqrt(disc))/(2*a);
               printf("x1=%f,x2=%f",x1,x2);
        }
}

当a==0时，如果b==0,就不能用-c/b
觉得你的思路不严密，不过，还是谢谢了

好！

没人帮忙吗？》？

  {if(disc>=0.0)
     {
     if(disc>0.0)
     x1=(-b+sqrt(disc))/(2*a);
     x2=(-b-sqrt(disc))/(2*a);
     printf("%f%f",x1,x2);
     else
     {x1=x2=-b/(2*a);
     printf("%f%f",x1,x2);}
     }
     }
     else
     printf("无解");
}


就是少了几个}}

#include<math.h>
void main()
{
    float a=0.0;
    float b=0.0;
    float c=0.0;
     float disc;
    double x1=0.000000,x2=0.000000;
    printf("input 3 numbers:");
    scanf("%f%f%f",&a,&b,&c);
    disc=b*b-4*a*c;
    if(a==0.0)
    if(b==0.0)
    printf("no answer due to input error\n");
    else printf("the single root is %f\n",-c/b);
    else
    {if(disc>=0.0)
     {
     if(disc>0.0)
     {x1=(-b+sqrt(disc))/(2*a);
     x2=(-b-sqrt(disc))/(2*a);
     printf("%f%f",x1,x2);
     }
     else
     {x1=x2=-b/(2*a);
     printf("%f%f",x1,x2);}
     }

     else
     {printf("无解");
     }
     getch();
    }

    }

问一下，最后的  getch( ) 有什么作用？

     ？？？？

起到暂停的作用！