Cubic Roots Solver Function in JS

gistfile1.js

function CubicSolve(a, b, c, d){

  b /= a;
  c /= a;
  d /= a;

  var discrim, q, r, dum1, s, t, term1, r13;

  q = (3.0*c - (b*b))/9.0;
  r = -(27.0*d) + b*(9.0*c - 2.0*(b*b));
  r /= 54.0;

  discrim = q*q*q + r*r;
  
  var roots = [ {real: 0, i: 0}, {real: 0, i: 0}, {real: 0, i: 0} ]
  
  term1 = (b/3.0);

  if (discrim > 0) { // one root real, two are complex
   s = r + Math.sqrt(discrim);
   s = ((s < 0) ? -Math.pow(-s, (1.0/3.0)) : Math.pow(s, (1.0/3.0)));
   t = r - Math.sqrt(discrim);
   t = ((t < 0) ? -Math.pow(-t, (1.0/3.0)) : Math.pow(t, (1.0/3.0)));
   
   roots[0].real = -term1 + s + t;
   term1 += (s + t)/2.0;
   roots[1].real = roots[2].real = -term1;
   term1 = Math.sqrt(3.0)*(-t + s)/2;
   
   roots[1].i = term1;
   roots[2].i = -term1;
   return roots;
  } // End if (discrim > 0)

  // The remaining options are all real
  

  if (discrim == 0){ // All roots real, at least two are equal.
   r13 = ((r < 0) ? -Math.pow(-r,(1.0/3.0)) : Math.pow(r,(1.0/3.0)));
   roots[0].real = -term1 + 2.0*r13;
   roots[2].real = roots[1].real = -(r13 + term1);
   return roots;
  } // End if (discrim == 0)

  // Only option left is that all roots are real and unequal (to get here, q < 0)
  q = -q;
  dum1 = q*q*q;
  dum1 = Math.acos(r/Math.sqrt(dum1));
  r13 = 2.0*Math.sqrt(q);
  
  roots[0].real = -term1 + r13*Math.cos(dum1/3.0);
  roots[1].real = -term1 + r13*Math.cos((dum1 + 2.0*Math.PI)/3.0);
  roots[2].real = -term1 + r13*Math.cos((dum1 + 4.0*Math.PI)/3.0);
  
  return roots;
} 

Yes that's right Math.pow doesn't work well for 1/3 for negative numbers.