#### Answer

$x=\left\{ -\dfrac{1}{2}, 3 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
2-\dfrac{5}{x}=\dfrac{3}{x^2}
,$ multiply both sides by the $LCD.$ Then express the resulting equation in the form $ax^2+bx+c=0.$ Use 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 $LCD$ of the denominators, $1,x,$ and $x^2$ is $x^2$ since it is the lowest expression which can be divided exactly by all the given denominators.
Multiplying both sides by the $LCD=
x^2
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
x^2\left( 2-\dfrac{5}{x} \right)=\left( \dfrac{3}{x^2} \right)x^2
\\\\
x^2(2)-x(5)=1(3)
\\\\
2x^2-5x=3
.\end{array}
In the form $ax^2+bx+c=0,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
2x^2-5x-3=0
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the factored form of the equation above is
\begin{array}{l}\require{cancel}
(2x+1)(x-3)=0
.\end{array}
Equating each factor to zero (Zero Product Property) results to
\begin{array}{l}\require{cancel}
2x+1=0
\\\\\text{OR}\\\\
x-3=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
2x+1=0
\\\\
2x=-1
\\\\
x=-\dfrac{1}{2}
\\\\\text{OR}\\\\
x-3=0
\\\\
x=3
.\end{array}
Upon checking, the solutions do not produce an expression with a $0$ in the denominator. Hence, the solutions are $
x=\left\{ -\dfrac{1}{2}, 3 \right\}
.$