$$(4y-3)(4y+3) = 0$$
Answer
$$ \begin{matrix}y_1 = \dfrac{ 3 }{ 4 } & y_2 = -\dfrac{ 3 }{ 4 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (4y-3)(4y+3) &= 0&& \text{simplify left side} \\[1 em]16y^2+12y-12y-9 &= 0&& \\[1 em]16y^2+12y-12y-9 &= 0&& \\[1 em]16y^2-9 &= 0&& \\[1 em] \end{aligned} $$
$ 16x^{2}-9 = 0 $ is a quadratic equation.
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