Answer
$x=\left\{ \dfrac{1-\sqrt{3}}{2},\dfrac{1+\sqrt{3}}{2} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
2x^2-2x=1
,$ 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}
2x^2-2x-1=0
.\end{array}
The quadratic equation above has $a=
2
, b=
-2
, c=
-1
.$ 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(2)(-1)}}{2(2)}
\\\\
x=\dfrac{2\pm\sqrt{4+8}}{4}
\\\\
x=\dfrac{2\pm\sqrt{12}}{4}
\\\\
x=\dfrac{2\pm\sqrt{4\cdot3}}{4}
\\\\
x=\dfrac{2\pm\sqrt{(2)^2\cdot3}}{4}
\\\\
x=\dfrac{2\pm2\sqrt{3}}{4}
\\\\
x=\dfrac{2(1\pm\sqrt{3})}{4}
\\\\
x=\dfrac{\cancel2(1\pm\sqrt{3})}{\cancel2(2)}
\\\\
x=\dfrac{1\pm\sqrt{3}}{2}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{1-\sqrt{3}}{2}
\\\\\text{OR}\\\\
x=\dfrac{1+\sqrt{3}}{2}
.\end{array}
Hence, $
x=\left\{ \dfrac{1-\sqrt{3}}{2},\dfrac{1+\sqrt{3}}{2} \right\}
.$
