$$(4b-7)(8b-8) = 0$$
Answer
$$ \begin{matrix}b_1 = 1 & b_2 = \dfrac{ 7 }{ 4 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (4b-7)(8b-8) &= 0&& \text{simplify left side} \\[1 em]32b^2-32b-56b+56 &= 0&& \\[1 em]32b^2-88b+56 &= 0&& \\[1 em] \end{aligned} $$
$ 32x^{2}-88x+56 = 0 $ is a quadratic equation.
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