The roots of polynomial $ p(y) $ are:
$$ \begin{aligned}y_1 &= -1+\sqrt{ 2 }i\\[1 em]y_2 &= -1-\sqrt{ 2 }i \end{aligned} $$The solutions of $ y^2+2y+3 = 0 $ are: $ y = -1+\sqrt{ 2 }i ~ \text{and} ~ y = -1-\sqrt{ 2 }i$.
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