C语言编程：求1元2次方程的根

Answer 1

#include
"stdio.h"
#include
"math.h"
void
main()
{
printf("求方程ax^2+bx+c=0的根：\n");
L:
printf("请分别输入a,b,c的值。\n");
double
a,b,c,x1,x2,x,y;
char
z;
printf("输入a，b，c=")；
scanf("a=%lf,b=%lf,c=%lf\n,&a,&b,&c");
printf("你所输入的方程为%lfx^2+%lfx+%lf=0。\n",a,b,c);
M:
printf("是否正确？（是y＼否n）\n");
scanf("%c",&z);
if
(z=='n')
goto
L;
else
if
(z=='y')
{
if
((b*b-4*a*c)>0)
x1=(-b+sqrt(b*b-4*a*c))/(2*a),
x2=(-b-sqrt(b*b-4*a*c))/(2*a),
printf("有两个不同的实根：\nx1=%lf\nx2=%lf\n",x1,x2),
printf("\n");
if
((b*b-4*a*c)==0)
x1=-b/(2*a),
printf("有两个相同的实根：\nx1=x2=%lf\n",x1),
printf("\n");
if
((b*b-4*a*c)<0)
x=-b/(2*a),
y=sqrt(4*a*c-b*b)/(2*a),
printf("有两表个不同的曾根：\nx1=%lf+%lfi\nx2=%lf-%lfi\n",x,y,x,y),
printf("\n");
goto
L
;
}
else
printf("请正确输入:\n");
goto
M;
}