How do you solve $2{{x}^{2}}+8x-3=0$ by completing the square?

Hint: In the problem they have mentioned to use the method of completing the square. We know that the method of completing the square is used to solve or find the roots of a quadratic equation. So first we will check whether the given equation is a quadratic equation or not. If the given equation is a quadratic equation and which is in the form of $a{{x}^{2}}+bx+c=0$, then we will go to the further steps. In the first step we will divide the given equation with the coefficient of ${{x}^{2}}$ to get the coefficient of ${{x}^{2}}$ as $1$. In the step two we will add and subtract the square of the half of the coefficient of $x$ in the above equation and rearrange the obtained equation, then we will apply the formula either ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ or ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ based the obtained equation in step two. In step three we will simplify and move the constants to another side of the equation. In step four we will apply square roots on both sides of the equation and simplify it to get the value of $x$.

Complete step by step answer:
Given equation, $2{{x}^{2}}+8x-3=0$
We can clearly see that the above equation is a quadratic equation, so we can use the method of completing the square to solve the given equation.
Comparing the given equation with $a{{x}^{2}}+bx+c=0$, then we will get
$a=2$, $b=8$, $c=-3$.
Dividing the given equation with $2$, then we will get
$\begin{align}
  & \dfrac{2{{x}^{2}}+8x-3}{2}=\dfrac{0}{2} \\
 & \Rightarrow \dfrac{2}{2}{{x}^{2}}+\dfrac{8}{2}x-\dfrac{3}{2}=0 \\
 & \Rightarrow {{x}^{2}}+4x-\dfrac{3}{2}=0 \\
\end{align}$
In the above equation the coefficient of $x$ is $4$. Adding the square of half of the coefficient of $x$ in the above equation, then we will get
$\begin{align}
  & {{x}^{2}}+4x-\dfrac{3}{2}+{{\left( \dfrac{1}{2}.4 \right)}^{2}}-{{\left( \dfrac{1}{2}.4 \right)}^{2}}=0 \\
 & \Rightarrow {{x}^{2}}+4x-\dfrac{3}{2}+{{\left( 2 \right)}^{2}}-{{\left( 2 \right)}^{2}}=0 \\
\end{align}$
Rearranging and rewriting the terms in the above equation, then we will get
$\Rightarrow {{x}^{2}}+2\left( 2 \right)\left( x \right)+{{\left( 2 \right)}^{2}}-\dfrac{3}{2}-4=0$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, then we will get
$\Rightarrow {{\left( x+2 \right)}^{2}}-\dfrac{3}{2}-4=0$
Taking the constants in the above equation to the other side of the equation, then we will get
$\begin{align}
  & \Rightarrow {{\left( x+2 \right)}^{2}}=\dfrac{3}{2}+4 \\
 & \Rightarrow {{\left( x+2 \right)}^{2}}=\dfrac{3+2\times 4}{2} \\
 & \Rightarrow {{\left( x+2 \right)}^{2}}=\dfrac{11}{2} \\
\end{align}$
Applying square root on both side of the above equation, then we will get
$\begin{align}
  & \Rightarrow \sqrt{{{\left( x+2 \right)}^{2}}}=\sqrt{\dfrac{11}{2}\times \dfrac{2}{2}} \\
 & \Rightarrow x+2=\pm \dfrac{\sqrt{22}}{2} \\
\end{align}$
From the above equation, we will get
$x+2=\dfrac{\sqrt{22}}{2}$ or $x+2=-\dfrac{\sqrt{22}}{2}$
Simplifying the above equations, then
$\begin{align}
  & x=\dfrac{\sqrt{22}}{2}-2 \\
 & \Rightarrow x=\dfrac{\sqrt{22}-2\times 2}{2} \\
 & \Rightarrow x=\dfrac{\sqrt{22}-4}{2} \\
\end{align}$ or $\begin{align}
  & x=-\dfrac{\sqrt{22}}{2}-2 \\
 & \Rightarrow x=\dfrac{-\sqrt{22}-2\times 2}{2} \\
 & \Rightarrow x=\dfrac{-\sqrt{22}-4}{2} \\
\end{align}$
$\therefore $ The solution for the given equation is $x=\dfrac{\sqrt{22}-4}{2}$ or $x=\dfrac{-\sqrt{22}-4}{2}$.

Note:
We can also calculate the solution for the equation by using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
Substituting the values of $a=2$, $b=8$, $c=-3$ in the above formula then we will get
$\begin{align}
  & x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow x=\dfrac{-8\pm \sqrt{{{8}^{2}}-4\left( 2 \right)\left( -3 \right)}}{2\left( 2 \right)} \\
 & \Rightarrow x=\dfrac{-8\pm \sqrt{64+24}}{4} \\
 & \Rightarrow x=\dfrac{-8\pm \sqrt{88}}{4} \\
 & \Rightarrow x=\dfrac{-8\pm 4\sqrt{22}}{4} \\
 & \Rightarrow x=-2\pm \sqrt{22} \\
\end{align}$
From both the methods we got the same result.