C实现解一元二次方程

#include <stdio.h>
#include <math.h>

int main(void)
{
	int a = 3;
	int b = 10;
	int c = 100;
	double data;
	double x1;//存放一个解
	double x2;//存放第二个
	data = c*c - 4*a*b;
	
	if(data>0)
	{
		x1 = (-b+sqrt(data))/(2*a);
		x2 = (-b-sqrt(data))/(2*a);
		printf("该一元二次方程有两个解\nx1 = %.2f\nx2 = %.2f\n",x1,x2);//打印两位小数　
	 } 
	 else if(data==0)
	 {
	 	x1 = (-b) / (2*a);
	 	x2 = x1;
	 	printf("该一元二次方程有唯一解\nx1=x2= %.2f\n",x1);//打印两位小数
	 }
	 else	printf("无解\n");
	 
	 return 0;
 }