$$(3y-4)(2y-4) = 0$$
Answer
$$ \begin{matrix}y_1 = 2 & y_2 = \dfrac{ 4 }{ 3 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (3y-4)(2y-4) &= 0&& \text{simplify left side} \\[1 em]6y^2-12y-8y+16 &= 0&& \\[1 em]6y^2-20y+16 &= 0&& \\[1 em] \end{aligned} $$
$ 6x^{2}-20x+16 = 0 $ is a quadratic equation.
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