Given that one solution for the equation $\dfrac{1}{2x^{2} + kx} = \dfrac{1}{2}$ is $x = -2$, find

1. Substitute the given root into the equation and solve for $k$.

Since $x=-2$ is a solution,
$\frac{1}{2(-2)^2+k(-2)}=\frac{1}{2}$
$2(-2)^2-2k=2$
$8-2k=2$
$-2k=-6$
$k=3$

1. Substitute $k=3$ into the original equation and solve for the other root.

Using $k=3$,
$\frac{1}{2x^2+3x}=\frac{1}{2}$
Cross-multiply:
$2=2x^2+3x$
$2x^2+3x-2=0$
Factorise:
$(2x-1)(x+2)=0$
So $x=\dfrac{1}{2}$ or $x=-2$.
Since $x=-2$ is the given solution, the second possible value is
$x=\dfrac{1}{2}.$