Answer
$x=\left\{ \dfrac{5-\sqrt{15}}{5},\dfrac{5+\sqrt{15}}{5} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
5x^2-10x+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{5x^2-10x+2}{5}=\dfrac{0}{5}
\\\\
x^2-2x+\dfrac{2}{5}=0
\\\\
x^2-2x=-\dfrac{2}{5}
.\end{array}
In the equation above, $b=
-2
.$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to
\begin{array}{l}\require{cancel}
\left( \dfrac{-2}{2} \right)^2
\\\\=
\left( -1 \right)^2
\\\\=
1.\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}
x^2-2x+1=-\dfrac{2}{5}+1
\\\\
x^2-2x+1=-\dfrac{2}{5}+\dfrac{5}{5}
\\\\
x^2-2x+1=\dfrac{3}{5}
.\end{array}
With the left side now a perfect square trinomial, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\left( x-1 \right)^2=\dfrac{3}{5}
.\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}
x-1=\pm\sqrt{\dfrac{3}{5}}
\\\\
x-1=\pm\sqrt{\dfrac{3}{5}\cdot\dfrac{5}{5}}
\\\\
x-1=\pm\sqrt{\dfrac{1}{25}\cdot15}
\\\\
x-1=\pm\dfrac{1}{5}\sqrt{15}
\\\\
x-1=\pm\dfrac{\sqrt{15}}{5}
\\\\
x=1\pm\dfrac{\sqrt{15}}{5}
\\\\
x=\dfrac{5}{5}\pm\dfrac{\sqrt{15}}{5}
\\\\
x=\dfrac{5\pm\sqrt{15}}{5}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{5-\sqrt{15}}{5}
\\\\\text{OR}\\\\
x=\dfrac{5+\sqrt{15}}{5}
.\end{array}
Hence, $
x=\left\{ \dfrac{5-\sqrt{15}}{5},\dfrac{5+\sqrt{15}}{5} \right\}
.$
