Unit 1.3 Bisection root of equation

// one more way to get the answer written in this code by comment where changes has to be made
#include <stdio.h>
#include <math.h>

#define f(x) ((x*x*x) - (2*x) - 5)
int main()
{
    float x0,x1,e,x2,y0,y1,y2;
    int i=0;
    printf("Enter the value of x0,x1,e");
    scanf("%f %f %f",&x0,&x1,&e);
    y0 = f(x0);
    y1 = f(x1);
    if(y1*y0>0){
        printf("\nIntial values are not correct");
        return 0;
    }
    while(fabs((x1-x0)/x1)>e){ // only do while
        x2=(x0+x1)/2;
        y2=f(x2);
        i++;

        if(y2*y0>0){
            x0 = x2;
            y0 = y2;
        }
        else{
            x1 = x2;
            y1 = y2;
        }
    } // condition (fabs(y2)>e);
    printf("\nHence the root is %f",x2);
    printf("\nTotal number of interations are %d",i);
    return 0;
}

/*
Ouput
Enter the value of x0,x1,e
-1 3 0.0001

Hence the root is 2.094604
Total number of iterations are 15
*/