How do you complete the square for $2{{x}^{2}}+8x$?

Hint: From the question we have been asked to complete the square for the equation $2{{x}^{2}}+8x$. We can solve the given question by completing square method by extracting the coefficient of ${{x}^{2}}$ as a factor and ten to that we will add and ultimately subtract the square of $\dfrac{1}{2}$ of the coefficient of $x$.

Complete step by step solution:
So, in the process of solving the question first we will extract the coefficient of ${{x}^{2}}$ which is 2 as a factor in the given equation. So, the equation will be reduced as follows.
$\Rightarrow 2{{x}^{2}}+8x$
$\Rightarrow 2\left( {{x}^{2}}+4x \right)$
Now, here half of the coefficient of variable $x$ in the equation will be as follows.
$\Rightarrow \dfrac{1}{2}\times 4=2$
So, the square of the half of the coefficient of $x$is will be ${{2}^{2}}=4$.
Here we will add and subtract the square of half the coefficient of $x$ which we found above to the equation. So, the equation will be reduced as follows.
$\Rightarrow 2\left( {{x}^{2}}+4x \right)$
$\Rightarrow 2\left( {{x}^{2}}+4x+{{2}^{2}}-4 \right)$
Here in the above expression we can see that inside the bracket we have a complete square of $x+2$.
So, the above equation will be reduced after rearranging the terms as follows.
$\Rightarrow 2\left( {{\left( x+2 \right)}^{2}}-4 \right)$
Therefore, we complete the square for the given question as follows.
 $\Rightarrow 2\left( {{\left( x+2 \right)}^{2}}-4 \right)$

Note: Students must be careful in doing the calculations. Here we should be very careful in understanding the question. Here we are asked to find the complete square so here we need not find the solution for x or roots for the equation. If we do that then our solution will be wrong.