Answer
$x=\left\{ \dfrac{-1-\sqrt{7}}{3},\dfrac{-1+\sqrt{7}}{3} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
2-2x=3x^2
,$ express in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
Using the properties of equality, in the form $ax^2+bx+c=0,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
-3x^2-2x+2=0
\\\\
\dfrac{-3x^2-2x+2}{-1}=\dfrac{0}{-1}
\\\\
3x^2+2x-2=0
.\end{array}
The quadratic equation above has $a=
3
, b=
2
, c=
-2
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
x=\dfrac{-2\pm\sqrt{2^2-4(3)(-2)}}{2(3)}
\\\\
x=\dfrac{-2\pm\sqrt{4+24}}{6}
\\\\
x=\dfrac{-2\pm\sqrt{28}}{6}
.\end{array}
Simplifying the radical by writing the radicand as an expression that contains a factor that is a perfect power of the index and then extracting the root of that factor result to
\begin{array}{l}\require{cancel}
x=\dfrac{-2\pm\sqrt{4\cdot7}}{6}
\\\\
x=\dfrac{-2\pm\sqrt{(2)^2\cdot7}}{6}
\\\\
x=\dfrac{-2\pm2\sqrt{7}}{6}
.\end{array}
Cancelling the common factors from all the terms results to
\begin{array}{l}\require{cancel}
x=\dfrac{\cancel2(-1)\pm\cancel2(1)\sqrt{7}}{\cancel2(3)}
\\\\
x=\dfrac{-1\pm\sqrt{7}}{3}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{-1-\sqrt{7}}{3}
\\\\\text{OR}\\\\
x=\dfrac{-1+\sqrt{7}}{3}
.\end{array}
Hence, $
x=\left\{ \dfrac{-1-\sqrt{7}}{3},\dfrac{-1+\sqrt{7}}{3} \right\}
.$
