The Quartic Formula

$$\def\sgn{\mathop{\rm sgn}} x = {-3b\pm\biggl(\sqrt{3\Bigl(3b^2-8ac+2a\root3\of{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}+2a\root3\of{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace-\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}\Bigr)}\pm\sqrt{3\Bigl(3b^2-8ac+2a\bigl({-1+\sqrt{-3}\over2}\bigr)\root3\of{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}+2a\bigl({-1-\sqrt{-3}\over2}\bigr)\root3\of{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace-\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}\Bigr)}\biggr)\pm\sgn\biggl(\Bigl(\sgn\bigl(-b^3+4abc-8a^2d\bigr)-{1\over2}\Bigr)\Bigl(\sgn\bigl(\max((2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3,\min(3b^2-8ac,3b^4+16a^2c^2+16a^2bd-16ab^2c-64a^3e))\bigr)-{1\over2}\Bigr)\biggr)\sqrt{3\Bigl(3b^2-8ac+2a\bigl({-1-\sqrt{-3}\over2}\bigr)\root3\of{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}+2a\bigl({-1+\sqrt{-3}\over2}\bigr)\root3\of{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace-\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}\Bigr)}\over12a}$$

The quartic formula gives the solutions of $ax^4+bx^3+cx^2+dx+e=0$ for real numbers $a$, $b$, $c$, $d$, $e$ with $a\neq0$.

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