#include <stdio.h>
#include <math.h>
int mian()
{   
      float a,b,c,x1,x2,z;   
      printf("请输入a,b,c,的值\n");   
      scanf("%f%f%f",&a,&b,&c);
      if(a==0)   
      {   
             if(b==0)            
                     printf("不是方程\n");        
             else         
                     printf("有一个解为x1=%.2f\n",-c/b);   
       }   
       else   
       {         
                z=b*b-4*a*c;   
                if(z>0)      
                        {                  
                                    x1=(-b-sqrt(z))/(2*a);
                                    x2=(-b+sqrt(z))/(2*a);              
                                     printf("有两根x1=%.2f\nx2=%.2f\n",x1,x2);      
                         }     
               else if(z<0)              
                                    {                    
                                              x1=-b/(2*a);            
                                                x2=sqrt(-z)/(2*a);      
                                                printf("x1=%.2f+%.2fi\n",x1,x2);      
                                                  printf("x1=%.2f-%.2fi\n",x1,x2);         
                                     }            
                      else               
                                    printf("有一个解为x1=%.2f\n",-b/(2*a));   
     }
    return 0;
}

我们地球人表示不知道什么是int mian()

改了一下

#include<stdio.h>
#include<math.h>
int main(void)
{  

      float a,b,c,x1,x2,z=0;        

      printf("请输入a,b,c,的值\n");  

       scanf("%f%f%f",&a,&b,&c);
      if(a==0)  

      {  

             if(b==0){
                 printf("不是方程\n");
             }else{
                 printf("有一个解为x1=%.2f\n",-c/b);
                 return 0;
             }        

       }  

       else{        

            z=b*b-4*a*c;
       }
      

       if(z>0)     

         {                 

             x1=(-b-sqrt(z))/(2*a);
             x2=(-b+sqrt(z))/(2*a);             

             printf("有两根x1=%.2f\nx2=%.2f\n",x1,x2);     

          }else if(z<0)             

          {                   

             x1=-b/(2*a);           

             x2=sqrt(-z)/(2*a);     

             printf("x1=%.2f+%.2f\n",x1,x2);     

             printf("x1=%.2f-%.2f\n",x1,x2);        

          }else{
              printf("有一个解为x1=%.2f\n",-b/(2*a));
          }   

    return 0;
}