The Quartic Formula

$x = \frac{- 3 b \pm (\sqrt{3 (3 b^{2} - 8 a c + 2 a \sqrt[3]{4 (2 c^{3} - 9 b c d + 27 a d^{2} + 27 b^{2} e - 72 a c e + \sqrt{(2 c^{3} - 9 b c d + 27 a d^{2} + 27 b^{2} e - 72 a c e)^{2} - 4 (c^{2} - 3 b d + 12 a e)^{3}})} + 2 a \sqrt[3]{4 (2 c^{3} - 9 b c d + 27 a d^{2} + 27 b^{2} e - 72 a c e - \sqrt{(2 c^{3} - 9 b c d + 27 a d^{2} + 27 b^{2} e - 72 a c e)^{2} - 4 (c^{2} - 3 b d + 12 a e)^{3}})})} \pm \sqrt{3 (3 b^{2} - 8 a c + 2 a (\frac{- 1 + \sqrt{- 3}}{2}) \sqrt[3]{4 (2 c^{3} - 9 b c d + 27 a d^{2} + 27 b^{2} e - 72 a c e + \sqrt{(2 c^{3} - 9 b c d + 27 a d^{2} + 27 b^{2} e - 72 a c e)^{2} - 4 (c^{2} - 3 b d + 12 a e)^{3}})} + 2 a (\frac{- 1 - \sqrt{- 3}}{2}) \sqrt[3]{4 (2 c^{3} - 9 b c d + 27 a d^{2} + 27 b^{2} e - 72 a c e - \sqrt{(2 c^{3} - 9 b c d + 27 a d^{2} + 27 b^{2} e - 72 a c e)^{2} - 4 (c^{2} - 3 b d + 12 a e)^{3}})})}) \pm sgn ((sgn (- b^{3} + 4 a b c - 8 a^{2} d) - \frac{1}{2}) (sgn (max ((2 c^{3} - 9 b c d + 27 a d^{2} + 27 b^{2} e - 72 a c e)^{2} - 4 (c^{2} - 3 b d + 12 a e)^{3}, min (3 b^{2} - 8 a c, 3 b^{4} + 16 a^{2} c^{2} + 16 a^{2} b d - 16 a b^{2} c - 64 a^{3} e))) - \frac{1}{2})) \sqrt{3 (3 b^{2} - 8 a c + 2 a (\frac{- 1 - \sqrt{- 3}}{2}) \sqrt[3]{4 (2 c^{3} - 9 b c d + 27 a d^{2} + 27 b^{2} e - 72 a c e + \sqrt{(2 c^{3} - 9 b c d + 27 a d^{2} + 27 b^{2} e - 72 a c e)^{2} - 4 (c^{2} - 3 b d + 12 a e)^{3}})} + 2 a (\frac{- 1 + \sqrt{- 3}}{2}) \sqrt[3]{4 (2 c^{3} - 9 b c d + 27 a d^{2} + 27 b^{2} e - 72 a c e - \sqrt{(2 c^{3} - 9 b c d + 27 a d^{2} + 27 b^{2} e - 72 a c e)^{2} - 4 (c^{2} - 3 b d + 12 a e)^{3}})})}}{12 a}$

The quartic formula gives the solutions of $a x^{4} + b x^{3} + c x^{2} + d x + e = 0$ for real numbers $a$ , $b$ , $c$ , $d$ , $e$ with $a \neq 0$ .

Directions: Choose all possibilities for the three $\pm$ signs with the last two equivalent. Use real cube roots if possible, and principal roots otherwise.