cubic formula

The three roots $r_1,r_2,r_3$ of a cubic polynomial equation $x^3+a{x}^2+bx+c=0$ are given by

$r_1$$=$$-{\displaystyle \frac{a}{3}}+{\left({\displaystyle \frac{-2a^3+9ab-27c+\sqrt{{(2a^3-9ab+27c)}^2+4{(-a^2+3b)}^3}}{54}}\right)}^{1/3}$

$+{\left({\displaystyle \frac{-2a^3+9ab-27c-\sqrt{{(2a^3-9ab+27c)}^2+4{(-a^2+3b)}^3}}{54}}\right)}^{1/3}$

$r_2$$=$$-{\displaystyle \frac{a}{3}}-{\displaystyle \frac{1+i\sqrt{3}}{2}}{\left({\displaystyle \frac{-2a^3+9ab-27c+\sqrt{{(2a^3-9ab+27c)}^2+4{(-a^2+3b)}^3}}{54}}\right)}^{1/3}$

$+{\displaystyle \frac{-1+i\sqrt{3}}{2}}{\left({\displaystyle \frac{-2a^3+9ab-27c-\sqrt{{(2a^3-9ab+27c)}^2+4{(-a^2+3b)}^3}}{54}}\right)}^{1/3}$

$r_3$$=$$-{\displaystyle \frac{a}{3}}+{\displaystyle \frac{-1+i\sqrt{3}}{2}}{\left({\displaystyle \frac{-2a^3+9ab-27c+\sqrt{{(2a^3-9ab+27c)}^2+4{(-a^2+3b)}^3}}{54}}\right)}^{1/3}$

$-{\displaystyle \frac{1+i\sqrt{3}}{2}}{\left({\displaystyle \frac{-2a^3+9ab-27c-\sqrt{{(2a^3-9ab+27c)}^2+4{(-a^2+3b)}^3}}{54}}\right)}^{1/3}$