自定义函数求一元二次方程

解题思路:注意判别式小于0的情况

注意事项:

参考代码:

#include<stdio.h>

#include<math.h>

void fun(float a,float b,float c){

float dirt;

dirt=b*b-4*a*c;

if(dirt>0){

printf("x1=%.3f x2=%.3f",(-b+sqrt(dirt))/2*a,(-b-sqrt(dirt))/2*a);

}

else if(dirt==0){

printf("x1=%.3f x2=%.3f",-b/2*a,-b/2*a);

}

else{

printf("x1=%.3f+%.3fi x2=%.3f-%.3fi",-b/(2*a),sqrt(-1*dirt)/(2*a),-b/(2*a),sqrt(-1*dirt)/(2*a));

}

}

int main()

{

    float a,b,c,i,dirt;

    scanf("%f %f %f",&a,&b,&c);

    fun(a,b,c);

    return 0;

}