$$(7-2y)\cdot(5+y) = 0$$
Answer
$$ \begin{matrix}y_1 = -5 & y_2 = \dfrac{ 7 }{ 2 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (7-2y)\cdot(5+y) &= 0&& \text{simplify left side} \\[1 em]35+7y-10y-2y^2 &= 0&& \\[1 em]-2y^2-3y+35 &= 0&& \\[1 em] \end{aligned} $$
$ -2x^{2}-3x+35 = 0 $ is a quadratic equation.
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