Answer
The solution is $x=-2$
Work Step by Step
$\dfrac{2x+1}{x-2}+\dfrac{3}{x}=\dfrac{-6}{x^{2}-2x}$
Take out common factor $x$ from the denominator of the fraction on the right side of the equation:
$\dfrac{2x+1}{x-2}+\dfrac{3}{x}=\dfrac{-6}{x(x-2)}$
Multiply the whole equation by $x(x-2)$:
$x(x-2)\Big[\dfrac{2x+1}{x-2}+\dfrac{3}{x}=\dfrac{-6}{x(x-2)}\Big]$
$(x)(2x+1)+3(x-2)=-6$
Evaluate the indicated operations:
$2x^{2}+x+3x-6=-6$
Take $6$ to the left side of the equation and simplify:
$2x^{2}+x+3x-6+6=0$
$2x^{2}+4x=0$
Take out common factor $2x$ from the left side:
$2x(x+2)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$2x=0$
$x=\dfrac{0}{2}$
$x=0$
$x+2=0$
$x=-2$
Since the original equation is undefined for $x=0$, the solution is just $x=-2$
