Starting from @Pace Nielsen answer $$w^3+5w^2+3w+J=0$$ The discriminant is $$\Delta =-(J+9) (27 J-13)$$
If $J\lt -9$, the real solution write $$w=\frac{1}{3} \left(-5+8 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(-\frac{27J+1151}{128} \right)\right)\right)$$$$w=\frac{1}{3} \left(-5+8 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(-\frac{27J+115}{128} \right)\right)\right)$$
For $-9 < J <0$, for $k=0,1,2$ $$w_k=\frac{1}{3} \left(-5+8 \sin\left( \frac{(4 k+1)\pi}{6}+\frac 13 \cos ^{-1}\left(\frac{27J+115}{128} \right)\right)\right)$$ whcih are much nicer than the radicals.