Question

5y^2-7y+1 find the zeros

Answer 1

$\huge\mathbb{A~N~S~W~E~R}$

$p(x) = 5 {y}^{2} - 7y + 1 \\ \\ 0 = 5 {y}^{2} - 7y + 1 \\ \\ - 1 = 5 {y}^{2} - 7y \\ \\ \frac{ - 1}{7 \times 5} = {y}^{2} - y \\ \\ \frac{ - 1}{ \: 35} = {y}$