How do you factor $2{{n}^{2}}-7n-4$?

Hint: For this problem we need to calculate the factors for the given quadratic equation. We know that the standard form of quadratic equation is $a{{x}^{2}}+bx+c$, so we will first compare the given equation with the standard form of the equation and write the values of $a$, $b$, $c$. Now we will split the middle term $bx$ as $bx={{b}_{1}}x+{{b}_{2}}x$, where ${{b}_{1}}{{b}_{2}}=ac$. So, we will calculate the value of $ac$ and from the factors of the value we will choose two factors as ${{b}_{1}}$, ${{b}_{2}}$. After that we will take appropriate terms as common from the equation and simplify it to get the factors of the given equation.

Complete step by step solution:
Given that, $2{{n}^{2}}-7n-4$.
We can observe that the above equation is a quadratic equation in terms of $n$. We have the standard form of quadratic equation in terms of $n$ as $a{{n}^{2}}+bn+c$. Comparing the given equation with the above equation, then we will get
$a=2$, $b=-7$, $c=-4$.
Now the value of $ac$ is $ac=2\left( -4 \right)=-8$. We have the factors of $8$ as $1$, $2$, $4$, $8$. From the above values, we can write
$\begin{align}
  & -1\times 8=-8 \\
 & -1+8=-7 \\
\end{align}$
So, we are going to split the term $-7n$ as $-n+8n$. Now the given equation is modified as
$\Rightarrow 2{{n}^{2}}-7n-4=2{{n}^{2}}-n+8n-4$
Taking $n$ as common from the first two terms and $4$ as common from the last two terms, then we will get
$\Rightarrow 2{{n}^{2}}-7n-4=n\left( 2n-1 \right)+4\left( 2n-1 \right)$
Again, taking $2n-1$ as common from the above equation, then we will have
$\Rightarrow 2{{n}^{2}}-7n-4=\left( 2n-1 \right)\left( n+4 \right)$
Hence the factors of the given equation $2{{n}^{2}}-7n-4$ are $2n-1$, $n+4$.

Note: The above procedure is a part of the completing squares method to find the roots of a quadratic equation. After getting the factors of the given equation we will equate each factor to zero individually and simplify the equations to get the roots.