please-i-need-cubic-formula-

Question Number 171829 by Mikenice last updated on 21/Jun/22

p l e a s e i n e e d c u b i c f o r m u l a

Answered by MathematicalUser2357 last updated on 05/Jan/24

x_{1} = - \frac{b}{3 a} - \frac{1}{3 a} \sqrt[3]{\frac{2 b^{3} - 9 a b c + 27 a^{2} d + \sqrt{{(2 b^{3} - 9 a b c + 27 a^{2} d)}^{2} - 4 {(b^{2} - 3 a c)}^{3}}}{2}} - \frac{1}{3 a} \sqrt[3]{\frac{2 b^{3} - 9 a b c + 27 a^{2} d - \sqrt{{(2 b^{3} - 9 a b c + 27 a^{2} d)}^{2} - 4 {(b^{2} - 3 a c)}^{3}}}{2}}

x_{2} = - \frac{b}{3 a} + \frac{1 + i \sqrt{3}}{6 a} \sqrt[3]{\frac{2 b^{3} - 9 a b c + 27 a^{2} d + \sqrt{{(2 b^{3} - 9 a b c + 27 a^{2} d)}^{2} - 4 {(b^{2} - 3 a c)}^{3}}}{2}} - \frac{1 - i \sqrt{3}}{6 a} \sqrt[3]{\frac{2 b^{3} - 9 a b c + 27 a^{2} d - \sqrt{{(2 b^{3} - 9 a b c + 27 a^{2} d)}^{2} - 4 {(b^{2} - 3 a c)}^{3}}}{2}}

x_{2} = - \frac{b}{3 a} - \frac{1 - i \sqrt{3}}{6 a} \sqrt[3]{\frac{2 b^{3} - 9 a b c + 27 a^{2} d + \sqrt{{(2 b^{3} - 9 a b c + 27 a^{2} d)}^{2} - 4 {(b^{2} - 3 a c)}^{3}}}{2} +} \frac{1 + i \sqrt{3}}{6 a} \sqrt[3]{\frac{2 b^{3} - 9 a b c + 27 a^{2} d - \sqrt{{(2 b^{3} - 9 a b c + 27 a^{2} d)}^{2} - 4 {(b^{2} - 3 a c)}^{3}}}{2}}