How do you find the zeros of y=2x^3-8x^2+18x-72?

This can be factored as $y = 2 {x}^{2} \left(x - 4\right) + 18 \left(x - 4\right) = \left(2 {x}^{2} + 18\right) \left(x - 4\right)$.
Setting as $y = 0$ and solving:
$0 = \left(2 {x}^{2} + 18\right) \left(x - 4\right)$
$x = \pm 3 i \mathmr{and} 4$