$$(5y-4)(3y-1) = 0$$
Answer
$$ \begin{matrix}y_1 = \dfrac{ 1 }{ 3 } & y_2 = \dfrac{ 4 }{ 5 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (5y-4)(3y-1) &= 0&& \text{simplify left side} \\[1 em]15y^2-5y-12y+4 &= 0&& \\[1 em]15y^2-17y+4 &= 0&& \\[1 em] \end{aligned} $$
$ 15x^{2}-17x+4 = 0 $ is a quadratic equation.
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