Answer
$\left\{\dfrac{2-\sqrt{2}}{2},\dfrac{2+\sqrt{2}}{2}\right\}$
Work Step by Step
Using $ax^2+bx+c=0,$ the given equation, $
2r^2-4r+1=0
,$ has
\begin{align*}
a=
2
,\text{ }b=
-4
,\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{-(-4)\pm\sqrt{(-4)^2-4(2)(1)}}{2(2)}
\\\\&=
\dfrac{4\pm\sqrt{16-8}}{4}
\\\\&=
\dfrac{4\pm\sqrt{8}}{4}
\\\\&=
\dfrac{4\pm\sqrt{4\cdot2}}{4}
\\\\&=
\dfrac{4\pm2\sqrt{2}}{4}
\\\\&=
\dfrac{\cancelto24\pm\cancelto12\sqrt{2}}{\cancelto24}
\\\\&=
\dfrac{2\pm\sqrt{2}}{2}
.\end{align*}
Hence, the solution set of the equation $
2r^2-4r+1=0
$ is $\left\{\dfrac{2-\sqrt{2}}{2},\dfrac{2+\sqrt{2}}{2}\right\}$.
