Answer
$x=${$2 - \sqrt {10},2 + \sqrt {10}$}
Work Step by Step
Given: $\dfrac{1}{x}+\dfrac{1}{x+2}=\dfrac{1}{3}$
Re-write the given equation as: $x^2-4x-6=0$
Factorize the expression with the help of quadratic formula. Quadratic formula suggests that $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$
This implies that $x=\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(-6)}}{2(1)}$
or, $x=\dfrac{4 \pm \sqrt {40}}{2}$
or, $x=2 - \sqrt {10},2 + \sqrt {10}$
Hence, our solution set is: $x=${$2 - \sqrt {10},2 + \sqrt {10}$}
