Answer
$\displaystyle \{\frac{5+\sqrt{73}}{2},\frac{5-\sqrt{73}}{2}\}.$
Work Step by Step
First, exclude any x that yields 0 in a denominator
$ x\neq 0,-3$
$\displaystyle \frac{1}{x}+\frac{1}{x+3}=\frac{1}{4} \qquad $ ... multiply with LCD = $4x(x+3)$
$ 4(x+3)+4x=x(x+3)\qquad $ ... write as $ ax^{2}+bx+c=0$
$4x+12+4x=x^{2}+3x $
$0=x^{2}-5x-12$
Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$ x=\displaystyle \frac{-(-5)\pm\sqrt{(-5)^{2}-4(1)(-12)}}{2(1)}$
$=\displaystyle \frac{5\pm\sqrt{25+48}}{2}=\frac{5\pm\sqrt{73}}{2}$
The solution set is
$\displaystyle \{\frac{5+\sqrt{73}}{2},\frac{5-\sqrt{73}}{2}\}.$
