Answer
$\left\{-\dfrac{5}{2},1\right\}$
Work Step by Step
Let $z=
(2n+1)
$. Then the given equation, $
12=(2n+1)^2+(2n+1)
,$ is equivalent to
\begin{align*}
12&=z^2+z
.\end{align*}
In the form $ax^2+bx+c=0$, the equation above is equivalent to
\begin{align*}
0&=z^2+z-12
\\
z^2+z-12&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z+4)(z-3)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving the variable, then
\begin{array}{l|r}
z+4=0 & z-3=0
\\
z=-4 & z=3
.\end{array}
Since $z=
(2n+1)
$, then by back-substitution,
\begin{array}{l|r}
2n+1=-4 & 2n+1=3
\\
2n=-4-1 & 2n=3-1
\\
2n=-5 & 2n=2
\\\\
n=-\dfrac{5}{2} & n=\dfrac{2}{2}
\\\\
& n=1
.\end{array}
Hence, the solution set of the equation $
(2n+1)
$ is $\left\{-\dfrac{5}{2},1\right\}$.
