Question

2y²-y-6=0 by factorisation
method
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Answer 1

p(y) = y² - 5y + 6

y² - 5y + 6 = 0

sum = -5

product = 6

no:s = -2,-3

y² - 2y - 3y + 6 = 0

y(y - 2) - 3 ( y - 2) = 0

(y - 2) (y - 3) = 0

y = 2

y = 3

Hope this helps you !

Answer 2

Answer:

$2y^2 - y - 6 = 0 \\ 2y ^{2} - 4y + 3y - 6 = 0 \\ 2y(y - 2) + 3(y - 2) = 0 \\ (2y + 3)(y - 2) = 0 \\ \\ y = \frac{ - 3}{2} or + 2$

Step-by-step explanation:

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