Answer
The solutions are $x=2\pm\sqrt{10}$
Work Step by Step
$\dfrac{1}{x}+\dfrac{1}{x+2}=\dfrac{1}{3}$
Multiply the whole equation by $3x(x+2)$:
$3x(x+2)\Big(\dfrac{1}{x}+\dfrac{1}{x+2}=\dfrac{1}{3}\Big)$
$3(x+2)+3x=x(x+2)$
$3x+6+3x=x^{2}+2x$
Take all terms to the right side and simplify:
$0=x^{2}+2x-3x-3x-6$
$x^{2}-4x-6=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
In this case, $a=1$, $b=-4$ and $c=-6$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-(-4)\pm\sqrt{(-4)^{2}-4(1)(-6)}}{2(1)}=\dfrac{4\pm\sqrt{16+24}}{2}=...$
$...=\dfrac{4\pm\sqrt{40}}{2}=\dfrac{4\pm2\sqrt{10}}{2}=2\pm\sqrt{10}$
