Answer
$\left\{\dfrac{1-\sqrt{5}}{4},\dfrac{1+\sqrt{5}}{4}\right\}$
Work Step by Step
In the form $ax^2+bx+c=0,$ the given equation, $
8x^2-4x=2
,$ is equivalent to
\begin{align*}
8x^2-4x-2&=0
\\\\
\dfrac{8x^2-4x-2}{2}&=\dfrac{0}{2}
\\\\
4x^2-2x-1&=0
.\end{align*}
The equation above has
\begin{align*}
a=
4
,\text{ }b=
-2
,\text{ and }c=
-1
.\end{align*}
Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{align*}\require{cancel}
x&=
\dfrac{-(-2)\pm\sqrt{(-2)^2-4(4)(-1)}}{2(4)}
\\\\&=
\dfrac{2\pm\sqrt{4+16}}{8}
\\\\&=
\dfrac{2\pm\sqrt{20}}{8}
\\\\&=
\dfrac{2\pm\sqrt{4\cdot5}}{8}
\\\\&=
\dfrac{2\pm\sqrt{4}\cdot\sqrt{5}}{8}
\\\\&=
\dfrac{2\pm2\sqrt{5}}{8}
\\\\&=
\dfrac{\cancelto12\pm\cancelto12\sqrt{5}}{\cancelto48}
\\\\&=
\dfrac{1\pm\sqrt{5}}{4}
.\end{align*}
Hence, the solution set of the equation $
8x^2-4x=2
$ is $\left\{\dfrac{1-\sqrt{5}}{4},\dfrac{1+\sqrt{5}}{4}\right\}$.
