题目：编程计算并输出一元二次方程ax '+bx+c=0的两个根

解析：先判断delta与0的关系，

        若delta>0,则有两个不同的实根，利用求根公式( -b 土  √b^2 -4*a*c)/2*a ；

        若delta=0,则有两个相同的实根，则根为x1 = x2 = -（b/2*a）;

        若delta<0,则无实根；

逻辑：if判读选择

# include <stdio.h>
# include <math.h>int main()
{double a = 0;double b = 0;double c = 0;double delta =0;double x1 = 0;double x2 = 0;int flag = 0;do{printf("输入a,b,c:\n");scanf("%lf %lf %lf", &a, &b, &c);delta = b * b - 4 * a * c;if (delta > 0){x1 = (-b + sqrt(delta)) / (2 * a);x2 = (-b - sqrt(delta)) / (2 * a);printf("有两个解，x1 = %f, x2 = %f\n", x1, x2);}else if (0 == delta){x1 = x2 = (-b) / (2 * a);printf("x1 = x2 = %f\n", x1, x2);}else{printf("无实数解!\n");}printf("继续 1，退出 0: ");scanf("%d", &flag);  //} while (flag);return 0;
}