# How do you find all the zeros of y=x^2-3x-4 with its multiplicities?

This quadratic has zeros $x = 4$ and $x = - 1$, both with multiplicity $1$.
$y = {x}^{2} - 3 x - 4 = \left(x - 4\right) \left(x + 1\right)$
Hence zeros $x = 4$ and $x = - 1$, both with multiplicity $1$.