# How do you find all zeros of y=x^2-2x-8?

Given $y = {x}^{2} - 2 x - 8$
${x}^{2} - 2 x - 8 = \left(x + 2\right) \left(x - 4\right)$
${x}^{2} - 2 x - 8 = 0$
when $x = - 2$ and when $x = 4$