How do you solve $2{{x}^{2}}-3x+1=0$ by completing the square?

Hint: To avoid arithmetic multiplying fraction multiply the given equation $2{{x}^{2}}-3x+1=0$ by $'8'$
Use the difference of square identity that is ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
For simplifying the problem replace $'x'$ terms by $'a'$ and $'b'$
Place $'x'$ term on the left side to find the value of $'x'$ and the coefficient on the right side.

Complete step by step solution:As per the given equation is $2{{x}^{2}}-3x+1=0$
Here to avoid arithmetic involving fraction multiply the whole equation by $'8'$
$8\left( 2{{x}^{2}}-3x+1 \right)=0\times 8$
So, after multiplying the modified equation will be.
$16{{x}^{2}}-24x+8=0$
Here you can write $'+9-1'$on the place of $'+8'$
$16{{x}^{2}}-24x+9-1=0$
Here in above equation, $16{{x}^{2}}-24x+9$
${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ formula
Here in $16{{x}^{2}}-24x+9$
$a=4x$
And $b=3$
Therefore, process this step:
${{\left( 4x-3 \right)}^{2}}-1=0$
Here, $'1'$ can be written as $'{{1}^{2}}'$ because ${{1}^{2}}=1$
So,
${{\left( 4x-3 \right)}^{2}}-{{1}^{2}}=0$
Now in above equation you can apply ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ identity.
Here, $a=4x-3$
$b=1$
Therefore,
$\left[ \left( 4x-3 \right)-1 \right]\left[ \left( 4x-3 \right)+1 \right]=0$
Opening, the bracket to solve the equation or simplifying the equation.
$\left[ 4x-3-1 \right]\left[ 4x-3+1 \right]=0$
$\left( 4x-4 \right)\left( 4x-2 \right)=0$
Choose common term from each equation in $\left( 4x-4 \right)$ $'4'$ is common term as it divides by both $'4x'$ and $'4'$ and in ${{2}^{nd}}$ equation $\left( 4x-2 \right)'2'$ is common term as it divides by $'4x'$ and $'2'$ both.
So,
$\left[ 4\left( x-1 \right) \right]\left[ 2\left( 2x-1 \right) \right]=0$
Multiply $4\times 2=8$
$8\left( x-1 \right)\left( 2x-1 \right)=0$
For finding the value of $x,$
$8\left( x-1 \right)=0$ or $8\left( 2x-1 \right)=0$
$8x-8=0$ or $16x-8=0$
$x=\dfrac{8}{8}$ or $x=\dfrac{8}{16}$
$x=1$ and $x=\dfrac{1}{2}$
The value of $x$ is $1$ and $\dfrac{1}{2}$

Additional Information:
In order to use the completing the square method, the value of $a$ in quadratic equation must be $1.$ If it is not $1,$ you will have to use AC method or the quadratic formula in order to solve for $x.$
Completing the square is a method used to solve quadratic equation $a{{x}^{2}}+bx+c$ where a must be $1.$ The goal is to force a perfect square trinomial on one side and then solve for $'x'$ by taking the square root of both sides.

Note:
Use the difference of square identity that is ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ make the equation into difference of square identity.
As you have multiplied the given equation by $'8'$ to avoid too much arithmetic involving fraction.
The digit $'8'$ is preferred to multiply because $8={{2.2}^{2}}$ the first factor of $2$makes the leading term into a perfect square. The additional ${{2}^{2}}$ factor avoids us having to divide, $3$ by $2$ and end up working with $\dfrac{1}{2}$ and $\dfrac{1}{4}$