Everybody knows the quadratic formula, which lets you solve this equation: a x² + b x + c = 0. Turns out it's not hard to solve when there's also an x³ term. There are either one or three solutions. This algorithm makes me happy.

use constant pi => 3.141592653589793;
sub cubic {
   # solve a cubic equation in the form
   # x^3 + a x^2 + b x + c = 0
   my ($a, $b, $c) = @_;
   my $q = $a*$a/9 - $b/3;
   my $r = ($a*$a/27 - $b/6)*$a + $c/2;
   my $s = $a / -3;
   my $d = $r*$r - $q*$q*$q;
   if ($d > 0) {
      my $t = (sqrt($d) + abs($r)) ** (1/3);
      my $u = ($t + $q / $t);
      return $r > 0 ? $s - $u : $s + $u;
   }
   else {
      my $t = atan2(sqrt(-$d), $r) / 3;
      my $u = 2 * sqrt($q); # $d <= 0 implies $q >= 0
      return (
         $s - $u * cos($t),
         $s - $u * cos($t + 2/3*pi),
         $s - $u * cos($t - 2/3*pi),
      );
   }
}
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