超好懂的求一元二次方程

解题思路:取三个函数来计算三种情况，再一个函数来判断b*b-4*a*b与0的关系进那种情况；

注意事项:了解如何计算二元一次方程的根，过程很简单；

参考代码:

#include<stdio.h>

#include<math.h>

double fun(double a,double b,double c)

{

double x,y;

x=(-b+sqrt(b*b-4*a*c))/(2*a);

y=(-b-sqrt(b*b-4*a*c))/(2*a);

printf("x1=%.3lf x2=%.3lf",x,y);

 } 

double fact(double a,double b,double c)

{

double h,q;

h=-b/(2*a);

q=-b/(2*a);

printf("x1=%.3lf x2=%.3lf",h,q);

}

double hard(double a,double b,double c)

{

double x,y,j,k;

x=-b/(2*a);

y=-b/(2*a);

j=sqrt(fabs(b*b-4*a*c))/(2*a);

k=sqrt(fabs(b*b-4*a*c))/(2*a);

printf("x1=%.3lf+%.3lfi x2=%.3lf-%.3lfi",x,j,y,k);

}

double heihei(double a,double b,double c)

{

if(b*b-4*a*c>0)

return fun(a,b,c);

else if(b*b-4*a*c==0)

return fact(a,b,c);

else

return hard(a,b,c);

}

int main()

{

double a,b,c;

scanf("%lf%lf%lf",&a,&b,&c);

heihei(a,b,c);

return 0;

}