please i need cubic formula


Question and Answers Forum

All Questions Topic List
Logic Questions
Previous in All Question Next in All Question
Previous in Logic Next in Logic
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

	$x_{1} = - \frac{b}{3 a} - \frac{1}{3 a} \sqrt[3]{\frac{2 b^{3} - 9 a b c + 27 a^{2} d + \sqrt{{(2 b^{3} - 9 a b c + 27 a^{2} d)}^{2} - 4 {(b^{2} - 3 a c)}^{3}}}{2}} - \frac{1}{3 a} \sqrt[3]{\frac{2 b^{3} - 9 a b c + 27 a^{2} d - \sqrt{{(2 b^{3} - 9 a b c + 27 a^{2} d)}^{2} - 4 {(b^{2} - 3 a c)}^{3}}}{2}}$ $x_{2} = - \frac{b}{3 a} + \frac{1 + i \sqrt{3}}{6 a} \sqrt[3]{\frac{2 b^{3} - 9 a b c + 27 a^{2} d + \sqrt{{(2 b^{3} - 9 a b c + 27 a^{2} d)}^{2} - 4 {(b^{2} - 3 a c)}^{3}}}{2}} - \frac{1 - i \sqrt{3}}{6 a} \sqrt[3]{\frac{2 b^{3} - 9 a b c + 27 a^{2} d - \sqrt{{(2 b^{3} - 9 a b c + 27 a^{2} d)}^{2} - 4 {(b^{2} - 3 a c)}^{3}}}{2}}$ $x_{2} = - \frac{b}{3 a} - \frac{1 - i \sqrt{3}}{6 a} \sqrt[3]{\frac{2 b^{3} - 9 a b c + 27 a^{2} d + \sqrt{{(2 b^{3} - 9 a b c + 27 a^{2} d)}^{2} - 4 {(b^{2} - 3 a c)}^{3}}}{2} +} \frac{1 + i \sqrt{3}}{6 a} \sqrt[3]{\frac{2 b^{3} - 9 a b c + 27 a^{2} d - \sqrt{{(2 b^{3} - 9 a b c + 27 a^{2} d)}^{2} - 4 {(b^{2} - 3 a c)}^{3}}}{2}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum