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10th grade math

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how do i solve for y in the equation -5y(6y - 6) - y = -4(y - 5)

  • 10th grade math -

    -5y(6y - 6) - y = -4(y - 5)
    -30y^2 + 30y - y = -4y + 20
    30y^2 - 33y + 20 = 0
    y = (22 ± √-1311)/60
    or
    (22 ± i √1311)/60 , two complex roots.

  • 10th grade math -

    how kan i solve this equation
    7a-3a+2a-a=16

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