C实现解一元二次方程

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#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;  }