#### Answer

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

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
(2x+1)(x-2)=-3
,$ use the FOIL Method and convert the given equation in the form $ax^2+bx+c=0.$ Then express the equation in factored form. Equate each factor to zero using the Zero Product Property. Finally, solve each resulting 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(-2)+1(x)+1(-2)=-3
\\\\
2x^2-4x+x-2=-3
\\\\
2x^2+(-4x+x)+(-2+3)=0
\\\\
2x^2-3x+1=0
.\end{array}
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $
2(1)=2
$ and the value of $b$ is $
-3
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
-1,-2
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
2x^2-x-2x+1=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)-(2x-1)=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x(2x-1)-(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-1)=0
.\end{array}
Equating each factor to zero (Zero Product Property), the solutions to the equation above are
\begin{array}{l}\require{cancel}
2x-1=0
\\\\\text{OR}\\\\
x-1=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
2x-1=0
\\\\
2x=1
\\\\
x=\dfrac{1}{2}
\\\\\text{OR}\\\\
x-1=0
\\\\
x=1
.\end{array}
Hence, $
x=\left\{ \dfrac{1}{2},1 \right\}
.$