$$(7+u)(4u-5) = 0$$
Answer
$$ \begin{matrix}u_1 = \dfrac{ 5 }{ 4 } & u_2 = -7 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (7+u)(4u-5) &= 0&& \text{simplify left side} \\[1 em]28u-35+4u^2-5u &= 0&& \\[1 em]4u^2+23u-35 &= 0&& \\[1 em] \end{aligned} $$
$ 4x^{2}+23x-35 = 0 $ is a quadratic equation.
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