Answer
$\displaystyle x=\frac{13\pm\sqrt{145}}{4}$
Work Step by Step
We are given:
$\frac{2x}{x-3}+\frac{1}{x}=4$
We multiply through by $x(x-3)$:
$x(x-3)(\frac{2x}{x-3}+\frac{1}{x})=x(x-3)*4$
$2x*x+(x-3)=4x^{2}-12x$
$2x^{2}+x-3=4x^{2}-12x$
$2x^{2}-13x+3=0$
We solve using the quadratic formula ($a=2,\ b=-13,\ c=3$):
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-13)\pm\sqrt{(-13)^{2}-4*2*3}}{2*2}$
$\displaystyle x=\frac{-(-13)\pm\sqrt{169-24}}{4}$
$\displaystyle x=\frac{13\pm\sqrt{145}}{4}$
