Answer
$\left\{ \dfrac{5}{7},1 \right\}$
Work Step by Step
The standard form of the given equation, $
7p(p-2)+2(p+4)=3
,$ is
\begin{array}{l}\require{cancel}
7p^2-14p+2p+8=3
\\\\
7p^2+(-14p+2p)+(8-3)=0
\\\\
7p^2-12p+5=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
7p^2-12p+5=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-12)\pm\sqrt{(-12)^2-4(7)(5)}}{2(7)}
\\\\=
\dfrac{12\pm\sqrt{144-140}}{14}
\\\\=
\dfrac{12\pm\sqrt{4}}{14}
\\\\=
\dfrac{12\pm\sqrt{(2)^2}}{14}
\\\\=
\dfrac{12\pm2}{14}
\\\\=
\dfrac{12-2}{14}
\text{ OR }
\dfrac{12+2}{14}
\\\\=
\dfrac{10}{14}
\text{ OR }
\dfrac{14}{14}
\\\\=
\dfrac{5}{7}
\text{ OR }
1
.\end{array}
Hence, the solutions are $
\left\{ \dfrac{5}{7},1 \right\}
.$
