Algebra 1

$d=-1$
Given:$\dfrac{d}{d+2}-\dfrac{2}{2-d}=\frac{d+6}{d^2-4}$ As we know $a^2-b^2=(a-b)(a+b)$ $\dfrac{d}{d+2}-\dfrac{2}{2-d}=\frac{d+6}{(d-2)(d+2)}$ Need to find least common denominator(LCD). $\dfrac{d}{d+2}+\dfrac{2}{d-2}=\dfrac{d+6}{(d-2)(d+2)}$ $d(d-2)+2)d+2)=d+6$ $d^2-d-2=0$ $(d+1)(d-2)=0$ Since, $d=2$ does not satisfy the given expression.Hence, the possible solution is $d=-1$