How do you solve $-2{{x}^{2}}-7x+4=0$ using completing the square?

Hint: For solving the given equation $-2{{x}^{2}}-7x+4=0$ using completing the square method, we first have to make the coefficient of ${{x}^{2}}$ equal to one. Since the coefficient of ${{x}^{2}}$ is equal to $-2$ we need to divide the given equation by $-2$ so as to obtain the quadratic equation ${{x}^{2}}+\dfrac{7}{2}x-2=0$. Then we have to multiply and divide the middle term by $2$ to get \[{{x}^{2}}+2x\left( \dfrac{7}{4} \right)-2=0\]. And then we have to ${{\left( \dfrac{7}{4} \right)}^{2}}$ on both the sides of the equation to obtain \[{{x}^{2}}+2x\left( \dfrac{7}{4} \right)+{{\left( \dfrac{7}{4} \right)}^{2}}-2={{\left( \dfrac{7}{4} \right)}^{2}}\]. Finally, with the help of the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ we can contract the term \[{{x}^{2}}+2x\left( \dfrac{7}{4} \right)+{{\left( \dfrac{7}{4} \right)}^{2}}\] as ${{\left( x+\dfrac{7}{4} \right)}^{2}}$ so that the equation can be solved by using basic algebraic operations.

Complete step by step solution:
The equation is given as
$\Rightarrow -2{{x}^{2}}-7x+4=0$
For using the completing the square method, we first need to make the coefficient of ${{x}^{2}}$ equal to one. So we divide both the sides by $-2$ to get
$\begin{align}
  & \Rightarrow \dfrac{1}{-2}\left( -2{{x}^{2}}-7x+4 \right)=\dfrac{0}{-2} \\
 & \Rightarrow {{x}^{2}}+\dfrac{7}{2}x-2=0 \\
\end{align}$
Now, we multiply and divide the middle term by $2$ to get
$\begin{align}
  & \Rightarrow {{x}^{2}}+2\times \dfrac{7}{2\times 2}x-2=0 \\
 & \Rightarrow {{x}^{2}}+2\times \dfrac{7}{4}x-2=0 \\
 & \Rightarrow {{x}^{2}}+2x\left( \dfrac{7}{4} \right)-2=0 \\
\end{align}$
Adding ${{\left( \dfrac{7}{4} \right)}^{2}}$ both the sides, we get
\[\Rightarrow {{x}^{2}}+2x\left( \dfrac{7}{4} \right)+{{\left( \dfrac{7}{4} \right)}^{2}}-2={{\left( \dfrac{7}{4} \right)}^{2}}\]
Now, from the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ we can write the above equation as
\[\Rightarrow {{\left( x+\dfrac{7}{4} \right)}^{2}}-2={{\left( \dfrac{7}{4} \right)}^{2}}\]
Adding $2$ both the sides, we get
\[\begin{align}
  & \Rightarrow {{\left( x+\dfrac{7}{4} \right)}^{2}}-2+2={{\left( \dfrac{7}{4} \right)}^{2}}+2 \\
 & \Rightarrow {{\left( x+\dfrac{7}{4} \right)}^{2}}={{\left( \dfrac{7}{4} \right)}^{2}}+2 \\
 & \Rightarrow {{\left( x+\dfrac{7}{4} \right)}^{2}}=\dfrac{49}{16}+2 \\
 & \Rightarrow {{\left( x+\dfrac{7}{4} \right)}^{2}}=\dfrac{81}{16} \\
\end{align}\]
Subtracting \[\dfrac{81}{16}\] from both sides
$\begin{align}
  & \Rightarrow {{\left( x+\dfrac{7}{4} \right)}^{2}}-\dfrac{81}{16}=0 \\
 & \Rightarrow {{\left( x+\dfrac{7}{4} \right)}^{2}}-{{\left( \dfrac{9}{4} \right)}^{2}}=0 \\
\end{align}$
Now, we know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. So the above equation can be written as
\[\begin{align}
  & \Rightarrow \left( x+\dfrac{7}{4}+\dfrac{9}{4} \right)\left( x+\dfrac{7}{4}-\dfrac{9}{4} \right)=0 \\
 & \Rightarrow \left( x+\dfrac{16}{4} \right)\left( x+\dfrac{2}{4} \right)=0 \\
 & \Rightarrow \left( x+4 \right)\left( x+\dfrac{1}{2} \right)=0 \\
 & \Rightarrow x=-4,x=-\dfrac{1}{2} \\
\end{align}\]

Note: Do not forget to make the coefficient of ${{x}^{2}}$ equal to one before beginning with the completing the square method. Otherwise the terms will become very much complicated when you will use the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ and the chances of mistakes will increase due to the increased calculations.