Answer
$x=\left\{ -\dfrac{15}{8},-1 \right\}
$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
4(2x+3)^2-(2x+3)-3=0
,$ use substitution and then the method on factoring trinomials. Then equate each factor to zero (Zero Product Property) and solve for the value of the variable. Finally use back-substitution and then solve for the value of the original variable.
$\bf{\text{Solution Details:}}$
Let $z=2x+3.$ Then the equation above is equivalent to
\begin{array}{l}\require{cancel}
4z^2-z-3=0
.\end{array}
To factor the trinomial expression above, note that the value of $ac$ is $
4(-3)=-12
$ and the value of $b$ is $
-1
.$
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 $\{
3,-4
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
4z^2+3z-4z-3=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}
(4z^2+3z)-(4z+3)=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
z(4z+3)-(4z+3)=0
.\end{array}
Factoring the $GCF=
(2z+3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(4z+3)(z-1)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then the solutions are
\begin{array}{l}\require{cancel}
4z+3=0
\\\\\text{OR}\\\\
z-1=0
.\end{array}
Since $z=2x+3,$ by back-substitution, the solutions are
\begin{array}{l}\require{cancel}
4(2x+3)+3=0
\\\\\text{OR}\\\\
(2x+3)-1=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
4(2x+3)+3=0
\\\\
8x+12+3=0
\\\\
8x+15=0
\\\\
8x=-15
\\\\
x=-\dfrac{15}{8}
\\\\\text{OR}\\\\
(2x+3)-1=0
\\\\
2x+3-1=0
\\\\
2x+2=0
\\\\
2x=-2
\\\\
x=-\dfrac{2}{2}
\\\\
x=-1
.\end{array}
Hence, the solutions are $
x=\left\{ -\dfrac{15}{8},-1 \right\}
.$