求ax^{2}+bx+c=0方程根（基础作业

#include <stdio.h>
#include<math.h>
int main()
{
   double a,b,c,disc,x1,x2,realpart,imagpart;
	//scanf("%lf,%lf,%lf",&a,&b,&c);
	a=1;b=2;c=2;
	printf("the equation:");
	if(fabs(a)<=1e-6)
		printf("this is not a equation\n");
	else
	{
	
		disc=b*b-4*a*c;
		if(fabs(disc)<=1e-6)
			printf("has two equal root:%8.4f\n",-b/(2*a));
		else
			if(disc>1e-6)
			{
			
				x1=(-b+sqrt(disc))/(2*a);
				x2=(-b-sqrt(disc))/(2*a);
				printf("has two real roots:%8.4f and %8.4f\n",x1,x2);
			}
		else
		{
		
			realpart=-b/(2*a);
			imagpart=sqrt(-disc)/(2*a);
			printf("has complex roots:\n");
			printf("%8.4f+%8.4fi\n",realpart,imagpart);
			printf("%8.4f-%8.4fi\n",realpart,imagpart);
		}
	}
   
   return 0;
}

1. 输入a,b,c,判断a值是否为0，为0方程不是二元一次方程；这里用1e-6即为很小的数，接近0
2. a不为0，判断disc是否为0，为0则有两个相同的根
3. 不为0 分为大于0和小于0两种
4. disc大于0则有两个不同实根
5. disc小于0则有两个共轭复根