The roots of polynomial $ p(d) $ are:
$$ \begin{aligned}d_1 &= 3+\sqrt{ 15 }i\\[1 em]d_2 &= 3-\sqrt{ 15 }i \end{aligned} $$The solutions of $ d^2-6d+24 = 0 $ are: $ d = 3+\sqrt{ 15 }i ~ \text{and} ~ d = 3-\sqrt{ 15 }i$.
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