Question

find the zeroes of the polynomial p(y)=y^2-1

Answer 1

y^2-1^2=0

;(y+1)(y-1)=0

;y=1,-1

is your answer

Answer 2

${y}^{2} - 1 = 0 \\ {y}^{2} = 1 \\ y = \binom{ + }{ - } \sqrt{1} \\ so \sqrt{1} and - \: \sqrt{1} are \: zeroes \: of \: polynomial$
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