$$4\cdot(4-d)\cdot(4+d) = 0$$
Answer
$$ \begin{matrix}d_1 = 4 & d_2 = -4 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 4\cdot(4-d)\cdot(4+d) &= 0&& \text{simplify left side} \\[1 em](16-4d)\cdot(4+d) &= 0&& \\[1 em]64+16d-16d-4d^2 &= 0&& \\[1 em]64+16d-16d-4d^2 &= 0&& \\[1 em]-4d^2+64 &= 0&& \\[1 em] \end{aligned} $$
$ -4x^{2}+64 = 0 $ is a quadratic equation.
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