$$\frac{18}{(d+3)(d-3)}+4 = \frac{d}{d+3}$$
Answer
$$ \begin{matrix}d_1 = 2 & d_2 = 3 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{18}{(d+3)(d-3)}+4 &= \frac{d}{d+3}&& \text{multiply ALL terms by } \color{blue}{ (d+3)(d-3) }. \\[1 em](d+3)(d-3)\cdot\frac{18}{(d+3)(d-3)}+(d+3)(d-3)\cdot4 &= (d+3)(d-3)\frac{d}{d+3}&& \text{cancel out the denominators} \\[1 em]18d^2-108d+162+4d^2-36 &= d^2-3d&& \text{simplify left side} \\[1 em]22d^2-108d+126 &= d^2-3d&& \text{move all terms to the left hand side } \\[1 em]22d^2-108d+126-d^2+3d &= 0&& \text{simplify left side} \\[1 em]21d^2-105d+126 &= 0&& \\[1 em] \end{aligned} $$
$ 21x^{2}-105x+126 = 0 $ is a quadratic equation.
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