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How do you solve the equation 3(y2+3) =28y?

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How do you solve the equation 3(y2+3) =28y?

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  1. 3(y2+3)=28y

    3x2y=6y+3x3

    6y+9=28y

    28-9=19

    19/6        

    ........._

    y=3.16

    just by doing that in my head i think that answer is correct mabey.


  2. => 3*2y+3*3=28y

    => 6y+9=28y

    => 28y-6y=9

    => 22y=9

    => y=9/22

    => y=0.409....
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