#include<stdio.h>
#include<math.h>
int main()
{
    double a,b,c,dis,p,q;
    printf("please intput a;b;c:");
    scanf("%lf%lf%lf",&a,&b,&c);
    dis=b*b-4*a*c;
    p=-b/(2*a);
    if(a!=0)
    {
   
        if(dis>=0)
        {
            
            q=sqrt(dis)/(2*a);
            if(dis==0)
                printf("root1=root2=%7.2lf\n",p);
            else
                printf("root1=%7.2lf\root2=%7.2lf\n",p+q,p-q);
            
        }
        else
        {
            dis=-dis;
            q=sqrt(dis)/(2*a);
            printf("root1=%7.2lf+%7.2lfi\troot2=%7.2lf-%7.2lfi",p,q,p,q);
        }
    }
    else
      printf("root=%7.2lf\n",-c/b);
      return 0;
        
}

那就百度一下 找个正确的代码 编译运行一下看看

https://bbs.bccn.net/search.php?searchid=90&orderby=lastpost&ascdesc=desc&searchsubmit=yes
这还只是本论坛C版中的搜索结果

我再也找不到更多的说辞了

已经自己改好了，可以正确运行了。谢谢各位大神
#include<stdio.h>
#include<math.h>
int main()
{
    double a,b,c,dis,p,q;
    printf("please intput a;b;c:");
    scanf("%lf%lf%lf",&a,&b,&c);
    dis=b*b-4*a*c;
    p=-b/(2*a);
    if(a!=0)
    {
   
        if(dis>=0)
        {
            
            q=sqrt(dis)/(2*a);
            if(dis==0)
                printf("root1=root2=%7.2lf\n",p);
            else
                printf("root1=%7.2f\t  root2=%7.2f\n",p+q,p-q);
            
        }
        else
        {
            dis=-dis;
            q=sqrt(dis)/(2*a);
            printf("root1=%7.2f+%7.2fi\t root2=%7.2f-%7.2fi",p,q,p,q);
        }
    }
    else
      printf("root=%7.2f\n",-c/b);
      return 0;
        
}

已经自己改好了，可以正确运行了。谢谢各位大神
#include<stdio.h>
#include<math.h>
int main()
{
    double a,b,c,dis,p,q;
    printf("please intput a;b;c:");
    scanf("%lf%lf%lf",&a,&b,&c);
    dis=b*b-4*a*c;
    p=-b/(2*a);
    if(a!=0)
    {
   
        if(dis>=0)
        {
            
            q=sqrt(dis)/(2*a);
            if(dis==0)
                printf("root1=root2=%7.2lf\n",p);
            else
                printf("root1=%7.2f\t  root2=%7.2f\n",p+q,p-q);
            
        }
        else
        {
            dis=-dis;
            q=sqrt(dis)/(2*a);
            printf("root1=%7.2f+%7.2fi\t root2=%7.2f-%7.2fi",p,q,p,q);
        }
    }
    else
      printf("root=%7.2f\n",-c/b);
      return 0;
        
}

太复杂了，而且特殊情况呢？试过么