Please factor 9y^2 + 12y − 5 = 0?

4 Answers

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  • 4 years ago

    9y^2 + 12y - 5 = 0

    ( 3y - 1)(3y + 5) = 0

    Hence

    3y - 1 = 0

    y = 1/3

    &

    3y + 5 = 0

    y = -5/3

  • Colin
    Lv 7
    4 years ago

    (3*x + 5)*(3*x - 1) = 0

    x = -5/3 or x = 1/3

  • DWRead
    Lv 7
    4 years ago

    9y² + 12y - 5 = 0

    Use quadratic formula

    y = [-12 ± √(12² – 4·9(-5))] / [2·9]

     = [-12 ± √324] / 18

     = [-12 ± 18] /18

     = -1⅔, ⅓

    9y² + 12y - 5 = 9(x+1⅔)(x-⅓) = (3x+5)(3x-1)

  • Dylan
    Lv 6
    4 years ago

    (3y-1)(3y+5) = 0

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