Answer
$\left\{\dfrac{2}{3}\pm\dfrac{\sqrt{2}}{3}i\right\}$
Work Step by Step
In the form $ax^2+bx+c=0,$ the given equation, $
3p^2=2(2p-1)
,$ is equivalent to
\begin{align*}\require{cancel}
3p^2&=2(2p)+2(-1)
&(\text{use Distributive Property}
\\
3p^2&=4p-2
\\
3p^2-4p+2&=0
.\end{align*}
The equation above has
\begin{align*}a=
3
,b=
-4
,\text{ and }c=
2
.\end{align*}
Using $
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$ or the Quadratic Formula, then
\begin{align*}
x&=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(3)(2)}}{2(3)}
\\\\&=
\dfrac{4\pm\sqrt{16-24}}{6}
\\\\&=
\dfrac{4\pm\sqrt{-8}}{6}
.\end{align*}
Using concepts of simplifying radicals, the equation above is equivalent to
\begin{align*}
x&=\dfrac{4\pm\sqrt{8\cdot(-1)}}{6}
\\\\&=
\dfrac{4\pm\sqrt{8}\cdot\sqrt{-1}}{6}
\\\\&=
\dfrac{4\pm\sqrt{4\cdot2}\cdot\sqrt{-1}}{6}
\\\\&=
\dfrac{4\pm\sqrt{4}\cdot\sqrt{2}\cdot\sqrt{-1}}{6}
\\\\&=
\dfrac{4\pm2\cdot\sqrt{2}\cdot i}{6}
&(\text{use }i=\sqrt{-1})
\\\\&=
\dfrac{\cancelto24\pm\cancelto12\cdot\sqrt{2}\cdot i}{\cancelto36}
\\\\&=
\dfrac{2\pm\sqrt{2}\cdot i}{3}
\\\\&=
\dfrac{2}{3}\pm\dfrac{\sqrt{2}}{3}i
.\end{align*}
Hence, the solution set of $
3p^2=2(2p-1)
$ is $
\left\{\dfrac{2}{3}\pm\dfrac{\sqrt{2}}{3}i\right\}
$.
