Answer
$\left\{\dfrac{3-i\sqrt{3}}{3},\dfrac{3+i\sqrt{3}}{3}\right\}$
Work Step by Step
In the form $ax^2+bx+c=0,$ the given equation, $
3p^2=6p-4
,$ is equivalent to
\begin{align*}
3p^2-6p+4&=0
.\end{align*}
The equation above has
\begin{align*}
a=
3
,\text{ }b=
-6
,\text{ and }c=
4
.\end{align*}
Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{align*}\require{cancel}
x&=
\dfrac{-(-6)\pm\sqrt{(-6)^2-4(3)(4)}}{2(3)}
\\\\&=
\dfrac{6\pm\sqrt{36-48}}{6}
\\\\&=
\dfrac{6\pm\sqrt{-12}}{6}
\\\\&=
\dfrac{6\pm\sqrt{12\cdot(-1)}}{6}
\\\\&=
\dfrac{6\pm\sqrt{12}\cdot\sqrt{-1}}{6}
\\\\&=
\dfrac{6\pm\sqrt{4\cdot3}\cdot\sqrt{-1}}{6}
\\\\&=
\dfrac{6\pm\sqrt{4}\cdot\sqrt{3}\cdot\sqrt{-1}}{6}
\\\\&=
\dfrac{6\pm2\cdot\sqrt{3}\cdot\sqrt{-1}}{6}
\\\\&=
\dfrac{6\pm2\cdot\sqrt{3}\cdot i}{6}
&(\text{use }i=\sqrt{-1})
\\\\&=
\dfrac{6\pm2i\sqrt{3}}{6}
\\\\&=
\dfrac{\cancelto36\pm\cancelto12i\sqrt{3}}{\cancelto36}
\\\\&=
\dfrac{3\pm i\sqrt{3}}{3}
.\end{align*}
Hence, the solution set of the equation $
z^2+z+1=0
$ is $\left\{\dfrac{3-i\sqrt{3}}{3},\dfrac{3+i\sqrt{3}}{3}\right\}$.
