Find $x$ in the equation $ax^3+bx^2+cx=d$

Solution 1:

It happens that equations of the form $ax^3+bx^2+cx+d=0$ and cubic polynomials in general, do have their own so-called discriminant $\Delta$. It is defined as: $$ \Delta=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd.\, $$ We have $3$ distinct cases:

• If $\Delta > 0$, then the equation has three distinct real roots.
• If $\Delta = 0$, then the equation has a multiple root and all its roots are real.
• If $\Delta < 0$, then the equation has one real root and two nonreal complex conjugate roots.

The solutions can then be obtained using the cubic formula: $$\begin{align}x &= \sqrt[\displaystyle3\,]{\left(\dfrac{-b^3}{27a^3} + \dfrac{bc}{6a^2} - \dfrac{d}{2a}\right) + \sqrt{\left(\dfrac{-b^3}{27a^3} + \dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 + \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)^3}}\\ & + \sqrt[\displaystyle3\,]{\left(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a}\right) - \sqrt{\left(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 + \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)^3}} - \dfrac{b}{3a}.\end{align}$$

I hope this helps.
Best wishes, $\mathcal H$akim.