#### Answer

$x=\left\{ -\dfrac{4}{3},5 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
(3x+1)(x-3)=2+3(x+5)
,$ 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}
3x(x)+3x(-3)+1(x)+1(-3)=2+3(x+5)
\\\\
3x^2-9x+x-3=2+3(x+5)
\\\\
3x^2-8x-3=2+3(x+5)
.\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}
3x^2-8x-3=2+3x+15
\\\\
3x^2+(-8x-3x)+(-3-2-15)=0
\\\\
3x^2-11x-20=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 $
3(-20)=-60
$ and the value of $b$ is $
-11
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-60\}, \{2,-30\}, \{3,-20\}, \{4,-15\}, \{5,-12\}, \{6,-10\},
\\
\{-1,60\}, \{-2,30\}, \{-3,20\}, \{-4,15\}, \{-5,12\}, \{-6,10\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
4,-15
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
3x^2+4x-15x-20=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}
(3x^2+4x)-(15x+20)=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x(3x+4)-5(3x+4)=0
.\end{array}
Factoring the $GCF=
(3x+4)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(3x+4)(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}
3x+4=0
\text{ OR }
x-5=0
.\end{array}
Solving each of the equations above results to
\begin{array}{l}\require{cancel}
3x+4=0
\\\\
3x=-4
\\\\
x=-\dfrac{4}{3}
\\\\\text{ OR }\\\\
x-5=0
\\\\
x=5
.\end{array}
Hence, the solutions are $
x=\left\{ -\dfrac{4}{3},5 \right\}
.$