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http://maths.ucd.ie/courses/math1200/algebra/algebranotes8-3.pdf

yes its in caculus.

You'll see why when you factor the left-hand side of the equation to
(z^2+2)*(z^3+6) = 0

\[(z^5+2z^3)+(6z^2+12)=z^3(z^2+2)+6(z^2+2)=(z^3+6)(z^2+2).\]

\(z^2+2=0\) is very easy to solve and you it gives the roots \(z=\pm \sqrt{2}i\).

You can use the demoivre theorem to solve \(z^3+6\), (i.e find the third roots of \(-6\)).

so this last term will give us three roots?

Yep, The notes vishal kothari gave are very good. You should probably read them.

put z=e^ix =cosx+isinx then proceed

You're welcome!