Método de Bisección en C, Usando funciones

# include<stdio.h>
# include<conio.h>
# include<math.h>
# include<process.h>
# include<string.h>
# define TOL 0.000001
# define F(x) (x)*log10(x) - 1.2

void Bisect();
int i=1, n;
float raiz = 1;

void main(){
 printf("\n Solucion por el Metodo de Biseccion \n");
 printf ("\n La ecuacion es: ");
 printf("\n\t\t\t (x)*log10x) -1.2 = 0 \n\n");
 printf("Numero Maximo de Iteracciones: ");
 scanf ("%d",&n);
 Bisect();
}
void Bisect(){
 float x1,x2,x0;
 float f0,f1,f2;
 int j=0;
  for (x2=1;;x2++){
   f2 = F(x2);
   if(f2 >0){
    break;
   }
   
  }
  for (x1=x2-1;;x2--){
   f1 = F(x1);
   if(f1<0){
    break;
   }
  }
printf ("\t\t------------------------------------\t\t");
printf ("\n\t\t Iteracion                 Raiz \n");
printf("\t\t-------------------------------------\t\t");
 for (;i<=n;i++){
  x0 = (x1+x2)/2.0;
  f0 = F(x0);
    if (f0==0){
     raiz = x0;
    }
  if(f0*f1<0){
   x2=x0;
  }
     else{
      x1 = x0;
      f1= f0;
     }
  printf("\n\t\t Iteracion %d ", i);
  printf("\t  :\t %f", x0);
  if(fabs((x1-x2)/x1) < TOL ){
  printf("\n\t\t ------------------------------------");
    printf("\n\t\t Raiz = %f",x0);
 printf("\n\t\t Iteracion = %d\n",i);
 printf("\t\t-------------------------------------");
 getch(); 
   exit(0);
   }
  }
printf("\n\t\t---------------------------------------");
printf("\n\t\t\t Raiz = %7.5f",x0);
printf("\n\t\t\t Iteracion = %d\n", i-1);
printf("\t\t-----------------------------------------");
 
getch();
}

Ejecutando: