(6y-4)(y-4)=7(y2-4y-12)

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вопрос

Expanding the left side:

(6y-4)(y-4) = 6y^2 - 24y - 4y + 16 = 6y^2 - 28y + 16

Expanding the right side:

7(y^2 - 4y - 12) = 7y^2 - 28y - 84

Now, the equation becomes:

6y^2 - 28y + 16 = 7y^2 - 28y - 84

Rearranging terms:

6y^2 - 7y^2 - 28y - 28y + 16 + 84 = 0
-y^2 - 56y + 100 = 0

This is a quadratic equation that can be solved using the quadratic formula.