Hi all,

please, can anyone help me with this equation:

3(Z^3) + 2(Z^2 ) + 6Z + 4 = 0

I switched to trigonometry representation using Moivere rule, but couldn't continue...

I came up with:

3r^3(cis 3a) + 2r^2(cis 2a) + 6r(cis a) = -4(cis 360k)

or if you want:

3r^3(cis a)^3 + 2r^2(cis a)^2 + 6r(cis a) = -4(cis 360k)

but what's now?

Thank you very much...