Let's call x=u+v, then you obtain (u+v)^3+A(u+v)+B=0 (A=-2E/C; B=2k/C)
then, solving some products etc.. you have: (u^3+v^3+B)+(u+v)(3uv+A)=0
put it into a system
1) u^3+v^3=-B
2) 3uv=-A (u+v cannot be=0)
We turn the second equation into u^3*v^3=-A^3/27
so we can solve a II° eq. since we have two number whose sum and product are known, then you take the III root and sum them, then you use Ruffini to lower the degree of the original equation, once you found one solution. Then you can easily solve the remaining II deg. eq.
Ok I made the thing a bit simple, there are some problems with logics, complex solutions etc.
If you want something more precise I suggest to search the net.
That is, in fact, the general reduced third degree equation. There is a standard formula, called "Cardano's formula". Maxos was leading you through it. I recommend you google on "Cardano's formua" or "Cubic formula" to see the whole thing.
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