3 years ago

I don't know what method I should use to solve for y in this equation: (y-3)^2=4y-12

1 answer:

sdas [7]3 years ago

5 0

(y-3)^2 -4y +12 = 0
[y^2 -6y +9] -4y +12 =0
y^2 -10y + 21 = 0
(y -3) (y-7) =0
y=3 ; y = 7

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