#### Answer

$x=\left\{ -6,\dfrac{1}{2} \right\}$

#### Work Step by Step

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