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14 June, 19:26

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

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  1. 14 June, 19:29
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    (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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