To find the root of a equation using Secant Method

Program:

#include<stdio.h>
#include<conio.h>
#include<math.h>
#define EPS 0.000001
#define MAXIT 50
#define F(x) (x) * (x) * (x) - 2 * (x) - 5
void main()
{
    float a, b, root, x1, x2;
    int count, status;
    clrscr();
    printf("Input two starting points : ");
    scanf("%f %f", &a, &b);
    sec(&a, &b, &x1, &x2, &root, &count, &status);
    if(status == 1)
    {
        printf("\nDivision by zero");
        printf("\nLast x1 = %f", x1);
        printf("\nLast x2 = %f", x2);
        printf("\nNo. of iterations = %d", count);
    }
    else if(status == 2)
    {
        printf("\nNo convergence in %d iterations", MAXIT);
    }
    else
    {
        printf("\nRoot = %f", root);
        printf("\n\nFunction value at root = %f", F(root));
        printf("\n\nNo. of iterations = %d", count);
        }
    getch();
}
sec(float *a, float *b, float *x1, float *x2, float *root,
int *count, int *status)
{
    float x3, f1, f2, error;
    *x1 = *a;
    *x2 = *b;
    f1 = F(*a);
    f2 = F(*b);
    *count = 1;
    begin:
    if(fabs(f1 - f2) <= 1.E-10)
    {
        *status = 1;
        return;
    }
    x3 = *x2 - f2 * (*x2 - *x1) / (f2 - f1);
    error = fabs((x3 - *x2) / x3);
    if(error > EPS)
    {
        if(*count == MAXIT)
        {
            *status = 2;
            return;
        }
        else
        {
            *x1 = *x2;
        }
        *x2 = x3;
        f1 = f2;
        f2 = F(x3);
        *count = *count + 1;
        goto begin;
    }
    else
    {
        *root = x3;
        *status = 3;
        return;
    }
}

Output:

Input two starting points : 2 3
Root = 2.094552
Function value at root = 0.000001
No. of iterations = 6