How do you solve $2{{x}^{2}}+12x-14=0$ by completing the square?

Hint: To solve the given quadratic equation $2{{x}^{2}}+12x-14=0$ by completing the square method we are going to first of all take 2 as common from the given equation. After that we are going to make the quadratic expression written inside the bracket as the perfect square. Perfect square means we have reduced the quadratic expression to ${{\left( ax+b \right)}^{2}}$.

Complete step by step answer:
The quadratic equation given in the above problem which we have to solve using completing the square as follows:
$2{{x}^{2}}+12x-14=0$
Now, taking 2 as common in the L.H.S of the above equation we get,
$2\left( {{x}^{2}}+6x-7 \right)=0$
Applying completing the square method by splitting $6x$ as $2.x.3$ in the above equation we get,
$2\left( {{x}^{2}}+2.x.3-7 \right)=0$
Now, if we add and subtract 9 in the bracket in the L.H.S of the above equation we get,
$\begin{align}
  & 2\left( {{x}^{2}}+2.x.3+9-9-7 \right)=0 \\
 & \Rightarrow 2\left( {{x}^{2}}+2.x.3+9-16 \right)=0 \\
\end{align}$
In the above equation, we can write ${{x}^{2}}+2.x.3+9={{\left( x+3 \right)}^{2}}$ then the above equation will look like:
$2\left( {{\left( x+3 \right)}^{2}}-16 \right)=0$
Dividing 2 on both the sides we get,
$\begin{align}
  & \dfrac{2\left( {{\left( x+3 \right)}^{2}}-16 \right)}{2}=\dfrac{0}{2} \\
 & \Rightarrow {{\left( x+3 \right)}^{2}}-16=0 \\
 & \Rightarrow {{\left( x+3 \right)}^{2}}-{{\left( 4 \right)}^{2}}=0 \\
\end{align}$
In the above equation, the L.H.S is of the form ${{a}^{2}}-{{b}^{2}}$ where $a=x+3,b=4$ so we can use the identity which is equal to:
${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
$\begin{align}
  & \left( x+3+4 \right)\left( x+3-4 \right)=0 \\
 & \Rightarrow \left( x+7 \right)\left( x-1 \right)=0 \\
\end{align}$
Equating each bracket in the above equation to 0 we get,
$\begin{align}
  & x+7=0 \\
 & \Rightarrow x=-7 \\
 & x-1=0 \\
 & \Rightarrow x=1 \\
\end{align}$

Hence, we got 2 solutions from the above solution i.e. 1 and -7.

Note: You can check the values of x that you are getting are correct or not by substituting these values in the given equation and see whether after putting these values, L.H.S and R.H.S of the equation is the same or not.
The values of x that we have solved above are:
$x=1,-7$
Substituting $x=1$ in the above equation we get,
$\begin{align}
  & 2{{\left( 1 \right)}^{2}}+12\left( 1 \right)-14=0 \\
 & \Rightarrow 2+12-14=0 \\
 & \Rightarrow 14-14=0 \\
 & \Rightarrow 0=0 \\
\end{align}$
As you can see that L.H.S = R.H.S in the above equation so the value of x as 1 is correct.
Similarly, you can check the other value of x also.