How to solve this equation in general $ax^3+bx^2+cx+d=0 ?$ 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 $ax^3+bx^2+cx+d=0 ?$

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 $$x={{{\sqrt[3]{4\left(-Q+\sqrt{4P^3+Q^2}\right)}}+{\sqrt[3]{4\left(-Q-\sqrt{4P^3‌​+Q^2}\right)}}-{2ab}}\over{6a}}.$$
