#### Answer

The solutions are $x=1$ and $x=\dfrac{3}{4}$

#### Work Step by Step

$2=\dfrac{3}{2x-1}+\dfrac{-1}{(2x-1)^{2}}$
Multiply the whole equation by $(2x-1)^{2}$:
$(2x-1)^{2}\Big[2=\dfrac{3}{2x-1}+\dfrac{-1}{(2x-1)^{2}}\Big]$
$2(2x-1)^{2}=3(2x-1)-1$
Evaluate the indicated operations:
$2(4x^{2}-4x+1)=6x-3-1$
$8x^{2}-8x+2=6x-3-1$
Take all terms to the left side and simplify:
$8x^{2}-8x+2-6x+3+1=0$
$8x^{2}-14x+6=0$
Divide the whole equation by $2$:
$\dfrac{1}{2}\Big(8x^{2}-14x+6=0\Big)$
$4x^{2}-7x+3=0$
Solve by factoring:
$(4x-3)(x-1)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$4x-3=0$
$4x=3$
$x=\dfrac{3}{4}$
$x-1=0$
$x=1$
The solutions are $x=1$ and $x=\dfrac{3}{4}$