#### Answer

$x=-\dfrac{7}{4}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\dfrac{x}{x+2}+\dfrac{1}{x}+3=\dfrac{2}{x^2+2x}
,$ factor all expressions that can be factored. Then multiply both sides by the $LCD.$ Use the concepts of solving quadratic equations to find the values of $x.$ Finally, do checking and ensure that any denominator does not become $0.$
$\bf{\text{Solution Details:}}$
The factored form of the equation above is
\begin{array}{l}\require{cancel}
\dfrac{x}{x+2}+\dfrac{1}{x}+3=\dfrac{2}{x(x+2)}
.\end{array}
The $LCD$ of the denominators, $x+2,x,1,$ and $x(x+2)$ is $x(x+2)$ since it is the lowest expression which can be divided exactly by all the given denominators.
Multiplying both sides by the $LCD=
x(x+2)
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
x(x+2)\left( \dfrac{x}{x+2}+\dfrac{1}{x}+3 \right)=\left(\dfrac{2}{x(x+2)} \right) x(x+2)
\\\\
x(x)+(x+2)(1)+x(x+2)(3)=1(2)
\\\\
x^2+x+2+3x^2+6x=2
\\\\
(x^2+3x^2)+(x+6x)+(2-2)=0
\\\\
4x^2+7x=0
.\end{array}
Factoring the $GCF$ results to
\begin{array}{l}\require{cancel}
x(4x+7)=0
.\end{array}
Equating each factor to zero (Zero Product Property) results to
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
4x+7=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
4x+7=0
\\\\
4x=-7
\\\\
x=-\dfrac{7}{4}
.\end{array}
If $x=0,$ the part of the given equation, $
\dfrac{1}{x}
,$ becomes $
\dfrac{1}{0}
,$ which is undefined. Hence, only $
x=-\dfrac{7}{4}
$ satisfies the given equation.