Find the root of equation ax^2+bx+c=0 (C language)

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One 、 subject

Find the equation ax²+bx+c=0 The root of the , use 3 When two functions are solved separately ：bx²-4ac Greater than 0、 be equal to 0 And less than 0 And output the result . Enter... From the main function a,b,c Value .

Two 、 Code and results

1. Code

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

float x1,x2,t,p,q;
int main()
{
    
    void Greater_than_Zero(float a,float b);
    void Equal_to_Zero(float a,float b);
    void Smaller_than_zero(float a,float b);
    float a,b,c;
    printf(" Please enter a,b,c Value :\n");
    scanf("%f,%f,%f",&a,&b,&c);
    printf(" solve equations :%4.2f*x*x+%4.2f*x+%4.2f=0\n",a,b,c);
    t=b*b-4*a*c;
    if(t>0)
    {
    
        Greater_than_Zero(a,b);
        printf("x1=%f\nx2=%f\n",x1,x2);
    }
    else if(t==0)
    {
    
        Equal_to_Zero(a,b);
        printf("x1=x2=%f\n",x1);
    }
    else
    {
    
        Smaller_than_zero(a,b);
        printf("x1=%f+%fi\nx2=%f-%fi\n",p,q,p,q);
    }
    return 0;
}

void Greater_than_Zero(float a,float b)
{
    
    x1=(-b+sqrt(t))/(2*a);
    x2=(-b+sqrt(t))/(2*a);
}

void Equal_to_Zero(float a,float b)
{
    
    x1=x2=(-b)/(2*a);
}

void Smaller_than_zero(float a,float b)
{
    
    p=-b/(2*a);
    q=sqrt(-t)/(2*a);
}

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2. Running results

（1） Two unequal real roots

（2） Two equal real roots

（3） The complex roots of two conjugates