# How do you find the zeros of y=x^3-4x^2+8?

One zero is $2$ the other two are $1 \pm \sqrt{5}$
$\left(x - 2\right) \left({x}^{2} - 2 x - 4\right)$ whose zeros are
$x = \frac{2 \pm \sqrt{4 + 4 \cdot 4}}{2} = 1 \pm \sqrt{5}$