How do you solve $3{{x}^{2}}-4x-2=0$ by completing the square?

Hint: Try to break the equation to square terms by reversing ${{\left( a+b \right)}^{2}}$ formula i.e. ${{\left( a \right)}^{2}}+2\cdot a\cdot b+{{\left( b \right)}^{2}}={{\left( a+b \right)}^{2}}$. After getting one square term try to convert the rest to another square term. Then apply ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to get the solutions of the given equation.

Complete step by step answer:
For solving this question, we have to find the values of ‘x’, by completing the square method as given by the question.
Completing square: we have to convert the whole equation to square terms by splitting as follows.
For the equation is ${{x}^{2}}+ax+c=0$, then splitting will be like $\begin{align}
  & {{x}^{2}}+ax+c=0 \\
 & \Rightarrow {{\left( x \right)}^{2}}+2\cdot x\cdot \dfrac{a}{2}+{{\left( \dfrac{a}{2} \right)}^{2}}-{{\left( \dfrac{a}{2} \right)}^{2}}+c=0 \\
 & \Rightarrow {{\left( x+\dfrac{a}{2} \right)}^{2}}-{{\left( \dfrac{a}{2} \right)}^{2}}+c=0 \\
\end{align}$
Now considering our equation,
$3{{x}^{2}}-4x-2=0$
before converting to compete square we have to reduce the coefficient of ‘${{x}^{2}}$’
Dividing by ‘3’ on both the sides we get,
$\begin{align}
  & \Rightarrow \dfrac{3{{x}^{2}}-4x-2}{3}=\dfrac{0}{3} \\
 & \Rightarrow {{x}^{2}}-\dfrac{4}{3}x-\dfrac{2}{3}=0 \\
\end{align}$
It can be written as
$\begin{align}
  & \Rightarrow {{\left( x \right)}^{2}}-2\cdot x\cdot \dfrac{4}{6}+{{\left( \dfrac{4}{6} \right)}^{2}}-{{\left( \dfrac{4}{6} \right)}^{2}}-\dfrac{2}{3}=0 \\
 & \Rightarrow {{\left( x-\dfrac{4}{6} \right)}^{2}}-\left( \dfrac{16}{36}+\dfrac{2}{3} \right)=0 \\
 & \Rightarrow {{\left( x-\dfrac{2}{3} \right)}^{2}}-\dfrac{\left( 16+24 \right)}{36}=0 \\
 & \Rightarrow {{\left( x-\dfrac{2}{3} \right)}^{2}}-\dfrac{40}{36}=0 \\
 & \Rightarrow {{\left( x-\dfrac{2}{3} \right)}^{2}}-\dfrac{10}{9}=0 \\
\end{align}$
As we know, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
So to bring our equation in ${{a}^{2}}-{{b}^{2}}$ form, $\dfrac{10}{9}$ can be written as $\dfrac{\sqrt{10}}{3}$
$\begin{align}
  & \Rightarrow {{\left( x-\dfrac{2}{3} \right)}^{2}}-{{\left( \dfrac{\sqrt{10}}{3} \right)}^{2}}=0 \\
 & \Rightarrow \left( x-\dfrac{2}{3}+\dfrac{\sqrt{10}}{3} \right)\cdot \left( x-\dfrac{2}{3}-\dfrac{\sqrt{10}}{3} \right)=0 \\
\end{align}$
Either
 $\begin{align}
  & \left( x-\dfrac{2}{3}+\dfrac{\sqrt{10}}{3} \right)=0 \\
 & \Rightarrow x=\dfrac{2}{3}-\dfrac{\sqrt{10}}{3} \\
\end{align}$
Or,
$\begin{align}
  & \left( x-\dfrac{2}{3}-\dfrac{\sqrt{10}}{3} \right)=0 \\
 & \Rightarrow x=\dfrac{2}{3}+\dfrac{\sqrt{10}}{3} \\
\end{align}$
These are the two solutions of the given equation.

Note:
Converting to square terms should be the first approach for solving completing square questions. We must have to get at least one square term containing ‘x’. The constant part can also be taken care as follows
From the solution part itself
$\begin{align}
  & {{\left( x-\dfrac{2}{3} \right)}^{2}}-\dfrac{10}{9}=0 \\
 & \Rightarrow {{\left( x-\dfrac{2}{3} \right)}^{2}}=\dfrac{10}{9} \\
\end{align}$
Taking square root both the sides, we get
$\begin{align}
  & \Rightarrow \sqrt{{{\left( x-\dfrac{2}{3} \right)}^{2}}}=\sqrt{\dfrac{10}{9}} \\
 & \Rightarrow \left( x-\dfrac{2}{3} \right)=\pm \dfrac{\sqrt{10}}{3} \\
\end{align}$
From where we can get the same values of ‘x’ as
Either $x=\dfrac{2}{3}-\dfrac{\sqrt{10}}{3}$
Or $x=\dfrac{2}{3}+\dfrac{\sqrt{10}}{3}$
This is another method for solving such questions.
Since it is a 2nd degree equation, we are getting two values of ‘x’ for which the equation is satisfied.