One solution of the equation $(k+1)x^2 + kx = 15$ is $x = -3$.

1. Use the fact that $x=-3$ satisfies the equation to form and solve an equation in $k$.

Substitute $x=-3$ into $(k+1)x^2+kx=15$:
$9(k+1)-3k=15$
$9k+9-3k=15$
$6k=6$
$k=1$

1. Substitute $k=-\dfrac{2}{5}$ and solve the resulting quadratic equation for the other root.

Using $k=-\dfrac{2}{5}$,
$(k+1)x^2+kx=15$
$\left(\frac{3}{5}\right)x^2-\frac{2}{5}x=15$
Multiply by $5$:
$3x^2-2x=75$
$3x^2-2x-75=0$
$(x+3)(3x-25)\ne 3x^2-2x-75$
Instead,
$3x^2-2x-75=0$
Since one root is $x=-3$, divide by $(x+3)$:
$3x^2-2x-75=(x+3)(3x-25)$
Thus the second root is
$3x-25=0$
$x=\frac{25}{3}$