кубическое уравнение

/* Cubic equation solution. Real coefficients case.
 
   int Cubic(double *x,double a,double b,double c);
   Parameters:
   x - solution array (size 3). On output:
       3 real roots -> then x is filled with them;
       1 real + 2 complex -> x[0] is real, x[1] is real part of 
                             complex roots, x[2] - non-negative 
                             imaginary part.
   a, b, c - coefficients, as described 
   Returns: 3 - 3 real roots;
            1 - 1 real root + 2 complex;
            2 - 1 real root + complex roots imaginary part is zero 
                (i.e. 2 real roots). 
*/
 
#include <math.h>   /* for sqrt(), fabs(), pow(), cos(), acos(). */
#define M_PI (3.141592653589793)
#define M_2PI (2.*M_PI)
 
int Cubic(double *x,double a,double b,double c) {
  double q,r,r2,q3;
  q=(a*a-3.*b)/9.; r=(a*(2.*a*a-9.*b)+27.*c)/54.;
  r2=r*r; q3=q*q*q;
  if(r2<q3) {
    double t=acos(r/sqrt(q3));
    a/=3.; q=-2.*sqrt(q);
    x[0]=q*cos(t/3.)-a;
    x[1]=q*cos((t+M_2PI)/3.)-a;
    x[2]=q*cos((t-M_2PI)/3.)-a;
    return(3);
  }
  else {
    double aa,bb;
    if(r<=0.) r=-r;
    aa=-pow(r+sqrt(r2-q3),1./3.); 
    if(aa!=0.) bb=q/aa;
    else bb=0.;
    a/=3.; q=aa+bb; r=aa-bb; 
    x[0]=q-a;
    x[1]=(-0.5)*q-a;
    x[2]=(sqrt(3.)*0.5)*fabs(r);
    if(x[2]==0.) return(2);
    return(1);
  }
}