#### Answer

$x=\left\{ \dfrac{5}{2}, 4 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\dfrac{13}{x^2+10}=\dfrac{2}{x}
,$ use cross-multiplication. Then use the concepts of solving quadratic equations to find the values of $x.$ Finally, do checking and ensure that any denominator does not become $0.$
$\bf{\text{Solution Details:}}$
Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to
\begin{array}{l}\require{cancel}
13(x)=(x^2+10)(2)
\\\\
13x=2x^2+20
.\end{array}
In the form $ax^2+bx+c=0,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
-2x^2+13x-20=0
\\\\
-1(-2x^2+13x-20)=-1(0)
\\\\
2x^2-13x+20=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-5)(x-4)=0
.\end{array}
Equating each factor to zero (Zero Product Property) results to
\begin{array}{l}\require{cancel}
2x-5=0
\\\\\text{OR}\\\\
x-4=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
2x-5=0
\\\\
2x=5
\\\\
x=\dfrac{5}{2}
\\\\\text{OR}\\\\
x-4=0
\\\\
x=4
.\end{array}
Upon checking, the solutions do not produce an expression with a $0$ in the denominator. Hence, the solutions are $
x=\left\{ \dfrac{5}{2}, 4 \right\}
.$