C语言求解一元二次方程

注意：
注意deta的判定条件。

/*-----------------------
功能：求解一元二次方程
无解时输出NAN
输入示例：
3
1 2.1 1
1 -2 1
1.3 1 1.2
输出示例：
-0.73 -1.37
1.00 1.00
NAN
-------------------------
Author: Zhang Kaizhou
Date: 2019-3-17 16:39:46
------------------------*/
#include 
#include 
#include 

typedef struct node{
    float a, b, c;
    struct node * pnext;
} Node;

void compute_res(Node * phead);
void list_tail_insert(Node ** pphead, Node ** pptail, Node * p);

int main(){
    Node * phead = NULL, * ptail = NULL;
    int n, i;
    scanf("%d", &n);
    for(i = 0; i < n; i++){
        Node * pnew = (Node *)calloc(1, sizeof(Node));
        scanf("%f %f %f", &pnew->a, &pnew->b, &pnew->c);
        list_tail_insert(&phead, &ptail, pnew);
    }
    compute_res(phead);

    return 0;
}

void compute_res(Node * phead){
    float deta, res1, res2;
    while(phead != NULL){
        deta = phead->b * phead->b - 4 * phead->a * phead->c;
        if(deta < 0){
            printf("NAN\n");
        }
        if(deta == 0){
            res1 = -phead->b / (2 * phead->a);
            res2 = -phead->b / (2 * phead->a);
            printf("%.2f %.2f\n", res1, res2);
        }
        if(deta > 0){
            res1 = (-phead->b + sqrt(deta)) / (2 * phead->a);
            res2 = (-phead->b - sqrt(deta)) / (2 * phead->a);
            printf("%.2f %.2f\n", res1, res2);
        }
        phead = phead->pnext;
    }
    return;
}

void list_tail_insert(Node ** pphead, Node ** pptail, Node * p){
    if(* pphead == NULL){
        * pphead = p;
        * pptail = p;
    }else{
        (* pptail)->pnext = p;
        * pptail = p;
    }
    return;
}