Answer
$x=\left\{ \dfrac{-1-\sqrt{7}}{3},\dfrac{-1+\sqrt{7}}{3} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
3r^2+2r-2=0
,$ use first the properties of equality to express the equation in the form $x^2\pm bx=c.$ Once in this form, complete the square by adding $\left( \dfrac{b}{2} \right)^2$ to both sides of the equal sign. Then express the left side as a square of a binomial while simplify the right side. Then take the square root of both sides (Square Root Property) and use the properties of equality to isolate the variable.
$\bf{\text{Solution Details:}}$
Using the properties of equality, in the form $x^2+bx=c,$ the given equation is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{3r^2+2r-2}{3}=\dfrac{0}{3}
\\\\
r^2+\dfrac{2}{3}r-\dfrac{2}{3}=0
\\\\
r^2+\dfrac{2}{3}r=\dfrac{2}{3}
.\end{array}
In the equation above, $b=
\dfrac{2}{3}
.$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to
\begin{array}{l}\require{cancel}
\left( \dfrac{\dfrac{2}{3}}{2} \right)^2
\\\\=
\left( \dfrac{2}{3}\div{2} \right)^2
\\\\=
\left( \dfrac{2}{3}\cdot\dfrac{1}{2} \right)^2
\\\\=
\left( \dfrac{1}{3} \right)^2
\\\\=
\dfrac{1}{9}
.\end{array}
Adding the value of $\left( \dfrac{b}{2} \right)^2,$ to both sides of the equation above results to
\begin{array}{l}\require{cancel}
r^2+\dfrac{2}{3}r+\dfrac{1}{9}=\dfrac{2}{3}+\dfrac{1}{9}
\\\\
r^2+\dfrac{2}{3}r+\dfrac{1}{9}=\dfrac{6}{9}+\dfrac{1}{9}
\\\\
r^2+\dfrac{2}{3}r+\dfrac{1}{9}=\dfrac{7}{9}
.\end{array}
With the left side now a perfect square trinomial, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\left( r+\dfrac{1}{3} \right)^2=\dfrac{7}{9}
.\end{array}
Taking the square root of both sides (Square Root Property), simplifying the radical and then isolating the variable, the equation above is equivalent to
\begin{array}{l}\require{cancel}
r+\dfrac{1}{3}=\pm\sqrt{\dfrac{7}{9}}
\\\\
r+\dfrac{1}{3}=\pm\sqrt{\dfrac{1}{9}\cdot7}
\\\\
r+\dfrac{1}{3}=\pm\dfrac{1}{3}\sqrt{7}
\\\\
r+\dfrac{1}{3}=\pm\dfrac{\sqrt{7}}{3}
\\\\
r=-\dfrac{1}{3}\pm\dfrac{\sqrt{7}}{3}
\\\\
r=\dfrac{-1\pm\sqrt{7}}{3}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
r=\dfrac{-1-\sqrt{7}}{3}
\\\\\text{OR}\\\\
r=\dfrac{-1+\sqrt{7}}{3}
.\end{array}
Hence, $
x=\left\{ \dfrac{-1-\sqrt{7}}{3},\dfrac{-1+\sqrt{7}}{3} \right\}
.$