Question 781799
It has a real solution of -2 but also has 4 complex solutions.

so using the euler equation I came up with 

z= |z|cos5θ + isin5θ

so -2 = 2 cos5θ + isin5θ
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z^5 + 32 --> 32cis(t), t = 0 + n*360, n = 0,1,2,3,4

Zeroes at 2cis(t/5)
--> 2, 2cis(72), 2cis(144), 2cis(216), 2cis(288)