#### Answer

The solutions are $\dfrac{-9- \sqrt{201}}{6}$ and $\dfrac{-9+ \sqrt{201}}{6}$.

#### Work Step by Step

Multiply the LCD $(x-2)(x+3)$ to both sides of the equation to get rid of the denominators:
\begin{array}{ccc}
&(x-2)(x+3)\left[\dfrac{2x}{x-2}+\dfrac{x}{x+3}\right] &= &-\dfrac{5}{x+3} \cdot (x-2)(x+3)
\\&(x-2)(x+3) \cdot \dfrac{2x}{x-2}+(x-2)(x+3) \cdot \dfrac{x}{x+3} &= &-\dfrac{5}{x+3} \cdot (x-2)(x+3)
\end{array}
Cancel simplify by cancelling common factors to obtain:
\begin{array}{ccc}
\require{cancel}
\\&\cancel{(x-2)}(x+3) \cdot \dfrac{2x}{\cancel{x-2}}+(x-2)\cancel{(x+3)} \cdot \dfrac{x}{\cancel{x+3}} &= &-\dfrac{5}{\cancel{x+3}} \cdot (x-2)\cancel{(x+3)}
\\&(x+3)(2x)+(x-2)(x) &= &-5(x-2)
\\&2x(x) + 2x(3) + x(x)-x(2) &= &-5(x) -(-5)(2)
\\&2x^2+6x+x^2-2x &= &-5x+10
\\&3x^2+4x &= &-5x+10
\end{array}
Put all terms on the left side. Note that when a term is transferred to the other side of an equation, it changes it sign to its opposite.
\begin{array}{ccc}
\\&3x^2+4x &= &-5x+10
\\&3x^2+4x+5x-10 &= &0
\\&3x^2+9x-10 &= &0
\end{array}
The trinomial cannot be factored.
Solve the equation using the quadratic formula with $a=3, b=9, c=-10$ to obtain:
$x=\dfrac{-9\pm \sqrt{9^2-4(3)(-10)}}{2(3)}
\\x=\dfrac{-9\pm\sqrt{81+120}}{6}
\\x=\dfrac{-9\pm \sqrt{201}}{6}$
Therefore, the solutions are $\dfrac{-9- \sqrt{201}}{6}$ and $\dfrac{-9+ \sqrt{201}}{6}$.