Answer
$$x=-\frac{3}{4}$$
Work Step by Step
$$\frac{4x}{x-2}+\frac{3}{x}=\frac{-6}{x(x-2)}$$
$$(x(x-2))(\frac{4x}{x-2}+\frac{3}{x})=(x(x-2))(\frac{-6}{x(x-2)})$$
$$4x^2+3(x-2)=-6$$
$$4x^2+3x-6=-6$$
$$4x^2+3x=0$$
$$x(4x+3)=0$$
$$x=0$$
0 is an extraneous solution, so $x\ne0$.
$$4x+3=0$$
$$x=-\frac{3}{4}$$
