#### Answer

$x=\left\{ -5,-\dfrac{1}{5} \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
(5x+1)(x+3)=-2(5x+1)
,$ in $ax^2+bx+c=0$ form. Then, express the equation in factored form. Next, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation.
$\bf{\text{Solution Details:}}$
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
5x(x)+5x(3)+1(x)+1(3)=-2(5x+1)
\\\\
5x^2+15x+x+3=-2(5x+1)
\\\\
5x^2+16x+3=-2(5x+1)
.\end{array}
Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
5x^2+16x+3=-10x-2
\\\\
5x^2+(16x+10x)+(3+2)=0
\\\\
5x^2+26x+5=0
.\end{array}
To factor the equation above, find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ In the expression above the value of $ac$ is $
5(5)=25
$ and the value of $b$ is $
26
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,25\}, \{5,5\},
\\
\{-1,-25\}, \{-5,-5\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
1,25
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
5x^2+1x+25x+5=0
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(5x^2+1x)+(25x+5)=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x(5x+1)+5(5x+1)=0
.\end{array}
Factoring the $GCF=
(5x+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(5x+1)(x+5)=0
.\end{array}
Equating each of the factors in the expression above to zero (Zero-Factor Theorem), then,
\begin{array}{l}\require{cancel}
5x+1=0
\text{ OR }
x+5=0
.\end{array}
Solving each of the equations above results to
\begin{array}{l}\require{cancel}
5x+1=0
\\\\
5x=-1
\\\\
x=-\dfrac{1}{5}
\\\\\text{ OR }\\\\
x+5=0
\\\\
x=-5
.\end{array}
Hence, the solutions are $
x=\left\{ -5,-\dfrac{1}{5} \right\}
.$