Question

3ysquare-2y-1 quadratic equation
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Answer 1

$3 {y}^{2} - 2y - 1$

$3 {y}^{2} - (3 - 1)y - 1$

$3 {y}^{2} - 3y + y - 1$

$3y(y - 1) + 1(y - 1)$

$(y - 1)(3y + 1)$

Answer 2

Answer:

3y^2 - 2y - 1 = 0 (Your question)

3y^2 -3y + y - 1 = 0

3y (y - 1) + 1 (y - 1) = 0

(3y + 1) (y - 1) = 0

(3y + 1) = 0 or (y - 1) = 0

3y = -1 or y = 1

y = -1/3 or y = 1 are the roots of the quadratic equation.

I hope you understand.

Thanks for asking