Answer
$\{2+\sqrt{10},2-\sqrt{10}\}.$
Work Step by Step
$\displaystyle \frac{1}{x}+\frac{1}{x+2}=\frac{1}{3};$
First, exclude any x that yields 0 in a denominator
$ x\neq 0,-2$
$\displaystyle \frac{1}{x}+\frac{1}{x+2}=\frac{1}{3} \qquad $ ... multiply with LCD = $3x(x+2)$
$ 3(x+2)+3x=x(x+2)\qquad $ ... write as $ ax^{2}+bx+c=0$
$3x+6+3x=x^{2}+2x $
$0=x^{2}-4x-6$
Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
$ x=\displaystyle \frac{-(-4)\pm\sqrt{(-4)^{2}-4(1)(-6)}}{2(1)}$
$=\displaystyle \frac{4\pm\sqrt{16+24}}{2}=\frac{4\pm\sqrt{40}}{2}=\frac{4\pm\sqrt{4(10)}}{2}$
$=\displaystyle \frac{4\pm 2\sqrt{10}}{2}=\frac{2(2\pm\sqrt{10})}{2}$$=2\pm\sqrt{10}$
The solution set is
$\{2+\sqrt{10},2-\sqrt{10}\}.$
