Answer
The solutions are $x=\dfrac{5}{2}\pm\dfrac{\sqrt{73}}{2}$
Work Step by Step
$\dfrac{1}{x}+\dfrac{1}{x+3}=\dfrac{1}{4}$
Multiply the whole equation by $4x(x+3)$:
$4x(x+3)\Big(\dfrac{1}{x}+\dfrac{1}{x+3}=\dfrac{1}{4}\Big)$
$4(x+3)+4x=x(x+3)$
$4x+12+4x=x^{2}+3x$
Take all terms to the right side of the equation:
$0=x^{2}+3x-4x-4x-12$
Simplify and rearrange:
$x^{2}-5x-12=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
In this case, $a=1$, $b=-5$ and $c=-12$.
Substitute the known values into the formula and evaluate:
$x=\dfrac{-(-5)\pm\sqrt{(-5)^{2}-4(1)(-12)}}{2(1)}=\dfrac{5\pm\sqrt{25+48}}{2}=...$
$...=\dfrac{5\pm\sqrt{73}}{2}=\dfrac{5}{2}\pm\dfrac{\sqrt{73}}{2}$
