下面求一元二次方程的根的程序错在哪里呢？能够运行，但结果是错误的。
#include "math.h"
float x1,x2,d,p,q;
getzero(a,b)
float a,b;
{x1=(-b+sqrt(d))/(2*a);x2=(-b-sqrt(d))/(2*a);}

eqzero(a,b)
float a,b;
{x1=x2=-b/(2*a);}

ltzero(a,b)
float a,b;
{p=-b/(2*a);q=sqrt(-d)/(2*a);}

main ()
{ float a,b,c,d;
printf ("input a,b,c:");
scanf ("%f%f%f",&a,&b,&c);
d=b*b-4*a*c;
printf ("方程的解为:");
if (d>0)
{getzero (a,b);printf ("x1=%5.2f\t x2=%5.2f\n",x1,x2);}
else if (d==0)
{eqzero(a,b);printf ("x1=x2=%5.2f\n",x1);}
else
{ltzero(a,b);printf ("x1=%5.2f+%5.2fi\t x2=%5.2f-%5.2fi\n",p,q,p,q);}
getch();
}

说实话……我看的……楼主写的看不懂……

求高人帮我解答下，谢谢了

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

double main (void)  {
   
    double a, b, c, x1, x2, d;

    printf("ax^2+bx+c=0\n");

    printf("please input a:");
    scanf("%lf", &a);

    printf("please input b:");
    scanf("%lf", &b);

    printf("please input c:");
    scanf("%lf", &c);

    d=sqrt(b*b-4*a*c);

     if (d>=0) {
        x1=(-b+d)/(2*a);
        x2=(-b-d)/(2*a);
        printf("%.2lf*x^2+%.2lfx+%.2lf=0\n", a, b, c);
        printf("x1=%.2lf\t x2=%.2lf \n \n", x1, x2);
        }
        else
        {
        printf(" The equation no solution\n");
        }

    return 0;
}
自己写的…你自己参考下……我看不懂你写的……悲剧……

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

double main (void)  {
   
    double a, b, c, x1, x2, d;

    printf("ax^2+bx+c=0\n");

    printf("please input a:");
    scanf("%lf", &a);

    printf("please input b:");
    scanf("%lf", &b);

    printf("please input c:");
    scanf("%lf", &c);

    d=sqrt(b*b-4*a*c);

     if (d>=0) {
        x1=(-b+d)/(2*a);
        x2=(-b-d)/(2*a);
        printf("%.2lf*x^2+%.2lfx+%.2lf=0\n", a, b, c);
        printf("x1=%.2lf\t x2=%.2lf \n \n", x1, x2);
        }
        else
        {
        printf(" The equation no solution\n");
        }

    return 0;
}
自己写的…你自己参考下……我看不懂你写的……悲剧……

手机信号不好发了两个帖子……还是opera浏览器发帖好用……不像uc…不能识别回车……这帖回的……省的老说我写程序结构乱……

这样写才规范，才正确；