求方程ax²+bx+c=0的根，用3个函数分别求当：b²-4ac大于 0、等于0 和 小于0时的根并输出结果。从主函数输入a,b,c的值。

#include<stdio.h>
#include<math.h>
float x1, x2,p,q;

int main()
{
	void greater(float a, float b, float asc);
	void equal(float a, float b, float asc);
	void lessthan(float a, float b, float asc);
	float a, b, c, asc;
	printf("please enter number for ( a,b,c )：");
	scanf_s("%f%f%f", &a, &b, &c);
	asc = b * b - 4 * a*c;
	if (asc>0)
	{
		greater(a, b, asc);
		printf("x1= %f\nx2= %f\n", x1, x2);
	}
	if (asc==0)
	{
		equal(a, b, asc);
		printf("x1 = x2= %f\n", x1);
	}
	if (asc<0)
	{
		lessthan(a, b, asc);
		printf("x1= %f+%f*i\nx2= %f-%f*i\n",p,q,p,q);
	}
	return 0;
}

void greater(float a, float b,float asc)
{
	x1 = (-b + sqrt(asc)) / (2 * a);
	x2 = (-b - sqrt(asc)) / (2 * a);
}
void equal(float a, float b, float asc)
{
	x1 = x2 = (-b) / (2 * a);
}
void lessthan(float a, float b, float asc)
{
	p = -b / (2 * a);
	q = sqrt(-asc) / (2 * a);
}