Answer
$ x=\dfrac{7+\sqrt{57}}{4}$ or $x=\dfrac{7-\sqrt{57}}{4}$
Work Step by Step
$\dfrac{3+\frac{1}{x}}{2-\frac{4}{x}}=x$
Simplify $\Rightarrow \dfrac{\frac{3x+1}{x}}{\frac{2x-4}{x}}=x$
Simplify $\Rightarrow \dfrac{3x+1}{2x-4}=x$
cross-multiply: $\Rightarrow 3x+1=x(2x-4)$
$\Rightarrow 3x+1=2x^2-4x$
$\Rightarrow 2x^2-7x-1=0$
Use the quadratic formula, where $a=2, b=-7$ and $c=-1$
$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-7)\pm\sqrt{(-7)^2-4(2)(-1)}}{2(2)}$
$\Rightarrow x=\dfrac{7\pm\sqrt{49+8}}{4}=\dfrac{7\pm\sqrt{57}}{4}$
$\Rightarrow x=\dfrac{7+\sqrt{57}}{4}$ or $x=\dfrac{7-\sqrt{57}}{4}$