Решение уравнений методом Лобачевского - нужна помощь

//////////////////////////////////////////////////////////////////////////////
//  Solving nonlinear equations (Lobachevsky method)
//  (c) Johna Smith, 1996
//
//  Method description:
//   Given: a0+a1x+a2x^2+...+anx^n=0
//   This method allows to find modulus of the greatest root of this equation
//   even if it's complex. But in last case there can appear several messages
//   about impossibilty of calculation root of negative number.
//   The main idea of this method is to change given equation to other
//   equation which roots equals to powered roots of given equation. For example
//   if roots of the given equation are x0,x1,.. xn then roots of new equation
//   will be x0^2, x1^2, ..., xn^2. Repeating this operation we get an equation
//   where one root is much greater than other ones. So we can easily
//   obtain modulus of the greatrest root of the given equation.
//   To obtain other roots of equation we need to divide given equation
//   by (x-x0) (where x0 is found root) and apply this method to result.
//
//////////////////////////////////////////////////////////////////////////////
#include <stdio.h>
#include <math.h>
#define N   4
#define N1  N+1
#define Iterations  15  // number of iterations
double a[N1]={24,-50,35,-10,1};
void main(void)
{
  double r,b[N1],c[N1],g,bi,d;
  int z,k;
  // printing given equation
  printf("%f",a[0]);
  for(int i=1;i<N1;i++) printf("%+fx^%d",a[i],i);
  printf("=0\n\n");
  // preparing auxiliary arrays b and c
  for (i=0;i<N1;i++)
  {
    b[i]=a[i]/a[N];
    c[i]=0;
  }
  // setting required parameters
  r=1/2.0;
  g=1;
  // make all iterations
  for(int y=0;y<Iterations;y++)
  {
    // calculate coefficients c[i] (coefficients of new equation)
    z=1;
    for(i=0;i<N1;i++)
    {
      bi=z*b[i];
      k=(i+1)/2;
      for(int j=i%2;j<N1;j+=2)
      {
        c[k]+=bi*b[j];
        k++;
      }
      z=-z;
    }
    d=z*c[N-1];
    // check whether we could calculate root of d
    if(d>0)
    {
      // calculating and printing new iteration
      g*=powl(d,r);
      printf("%f\n",g);
      for (i=0;i<N1;i++)
      {
        // preparing data for next iteration
        b[i]=c[i]/powl(d,N-i);
        c[i]=0;
      }
      b[N-1]=z;
      b[N]=-z;
    } else
    {
      // d is negative - can't calculate root
      for(i=0;i<N1;i++)
      {
        // preparing data for next iteration
        b[i]=c[i];
        c[i]=0;
      }
      printf("no iteration (can't calculate root from negative number)\n");
    }
    r/=2.0;
  }
}