Answer
$\left\{-2,\dfrac{3}{2}\right\}$
Work Step by Step
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{
equation
}$
\begin{array}{l}\require{cancel}
2x^2+x-6=0
\end{array} has $ac=
2(-6)=-12
$ and $b=
1
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
-3,4
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
2x^2-3x+4x-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^2-3x)+(4x-6)=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x(2x-3)+2(2x-3)=0
.\end{array}
Factoring the $GCF=
(2x-3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(2x-3)(x+2)=0
.\end{array}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
2x-3=0 & x+2=0
\\
2x=3 & x=-2
\\\\
x=\dfrac{3}{2}
\end{array}
Hence, the solution set of the equation $
2x^2+x-6=0
$ is $\left\{-2,\dfrac{3}{2}\right\}$.
