C语言--二元一次方程求解(完)

(10)求方程式ax^2+bx+c=0的根.

#include<stdio.h>#include<math.h>int main(){int a,b,c;printf("请输入:");scanf("%d%d%d",&a,&b,&c);double delta,x1,x2,m,n;delta=b*b-4*a*c;if(delta>0){x1=(-b+sqrt(delta))/(2*a);x2=(-b-sqrt(delta))/(2*a);printf("原方程的的解为x1=%f,x2=%f\n",x1,x2);}if(delta==0){x1=(-b)/(2*a);x2=x1;printf("原方程的的解为x1=%f,x2=%f\n",x1,x2);}if(delta<0){m=(-b)/(2*a);n=sqrt(-delta)/(2*a);printf("原方程的的解为:x1=%8.4f+%8.4fi ; x2=%8.4f-%8.4fi\n ",m,n,m,n);}return 0;}