Answer
$\left\{-\dfrac{4}{3},\dfrac{1}{2}\right\}$
Work Step by Step
Let $z=
\dfrac{1}{2p+2}
$. Then the given equation, $
3-\dfrac{7}{2p+2}=\dfrac{6}{(2p+2)^2}
,$ is equivalent to
\begin{align*}\require{cancel}
3-7\cdot\dfrac{1}{2p+2}&=6\cdot\dfrac{1}{(2p+2)^2}
\\\\
3-7z&=6z^2
\\
0&=6z^2+7z-3
\\
6z^2+7z-3&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(3z-1)(2z+3)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
3z-1=0 & 2z+3=0
\\
3z=1 & 2z=-3
\\\\
z=\dfrac{1}{3} & z=-\dfrac{3}{2}
.\end{array}
Since $z=
\dfrac{1}{2p+2}
,$ by back substitution, then
\begin{array}{l|r}
\dfrac{1}{2p+2}=\dfrac{1}{3} & \dfrac{1}{2p+2}=-\dfrac{3}{2}
.\end{array}
By cross-multiplication and the properties of equality, the equations above are equivalent to
\begin{array}{l|r}
1(3)=1(2p+2) & 1(2)=(2p+2)(-3)
\\
3=2p+2 & 2=-6p-6
\\
3-2=2p & 6p=-6-2
\\
1=2p & 6p=-8
\\\\
\dfrac{1}{2}=p & p=-\dfrac{8}{6}
\\\\
p=\dfrac{1}{2} & p=-\dfrac{4}{3}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }p=\dfrac{1}{2}: & \text{If }p=-\dfrac{4}{3}
\\\\
3-\dfrac{7}{2\left(\frac{1}{2}\right)+2}\overset{?}=\dfrac{6}{\left(2\left(\frac{1}{2}\right)+2\right)^2} &
3-\dfrac{7}{2\left(-\frac{4}{3}\right)+2}\overset{?}=\dfrac{6}{\left(2\left(-\frac{4}{3}\right)+2\right)^2}
\\\\
3-\dfrac{7}{1+2}\overset{?}=\dfrac{6}{\left(1+2\right)^2} &
3-\dfrac{7}{-\frac{8}{3}+2}\overset{?}=\dfrac{6}{\left(-\frac{8}{3}+2\right)^2}
\\\\
3-\dfrac{7}{3}\overset{?}=\dfrac{6}{\left(3\right)^2} &
3-\dfrac{7}{-\frac{8}{3}+\frac{6}{3}}\overset{?}=\dfrac{6}{\left(-\frac{8}{3}+\frac{6}{3}\right)^2}
\\\\
\dfrac{9}{3}-\dfrac{7}{3}\overset{?}=\dfrac{6}{9}&
3-\dfrac{7}{-\frac{2}{3}}\overset{?}=\dfrac{6}{\left(-\frac{2}{3}\right)^2}
\\\\
\dfrac{2}{3}\overset{?}=\dfrac{\cancelto26}{\cancelto39}&
3+\dfrac{21}{2}\overset{?}=\dfrac{6}{\frac{4}{9}}
\\\\
\dfrac{2}{3}\overset{\checkmark}=\dfrac{2}{3}&
\dfrac{6}{2}+\dfrac{21}{2}\overset{?}=\dfrac{54}{4}
\\\\
&
\dfrac{27}{2}\overset{?}=\dfrac{\cancelto{27}{54}}{\cancelto24}
\\\\
&
\dfrac{27}{2}\overset{\checkmark}=\dfrac{27}{2}
.\end{array}
Since all solutions satisfy the original equation, then the solution set of the equation $
3-\dfrac{7}{2p+2}=\dfrac{6}{(2p+2)^2}
$ is $\left\{-\dfrac{4}{3},\dfrac{1}{2}\right\}$.
