For example if you have $ax^2+bx+c=0$ then solution given by formula $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
Is there formula like this for $ax^3+bx^2+cx+d=0 ?$
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http://en.wikipedia.org/wiki/Cubic_function – Claude Leibovici Jun 19 '14 at 18:19
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Yes, such formula exists, and it is called the cubic formula, however it isn't nice looking: $$\begin{align}x &= \sqrt[\displaystyle3\,]{\left(\dfrac{-b^3}{27a^3} + \dfrac{bc}{6a^2} - \dfrac{d}{2a}\right) + \sqrt{\left(\dfrac{-b^3}{27a^3} + \dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 + \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)^3}}\\ & + \sqrt[\displaystyle3\,]{\left(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a}\right) - \sqrt{\left(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 + \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)^3}} - \dfrac{b}{3a}.\end{align}$$ It can be made more simple, as suggested by @SenexÆgyptiParvi, if you let $P=3ac-b^2$ and $Q=27a^2d-9abc+2b^3$, then $$x={{{\sqrt[3]{4\left(-Q+\sqrt{4P^3+Q^2}\right)}}+{\sqrt[3]{4\left(-Q-\sqrt{4P^3+Q^2}\right)}}-{2ab}}\over{6a}}.$$
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I broke it into two lines, which confirmed my suspicion that your two cube roots are identical. Should there be a minus sign in front of one of the square roots inside? – Harald Hanche-Olsen Jun 19 '14 at 18:22
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1A bit prettier, perhaps: If $P=3ac-b^2$ and $Q=27a^2d-9abc+2b^3$, then $x={{{\sqrt[3]{4\left(-Q+\sqrt{4P^3+Q^2}\right)}}+{\sqrt[3]{4\left(-Q-\sqrt{4P^3+Q^2}\right)}}-{2ab}}\over{6a}}$ – Senex Ægypti Parvi Jun 19 '14 at 22:18
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