How do you factor $42{{n}^{2}}-n-30$ ?

Hint: In this question, we have to find the factors of the given quadratic equation. we will solve it by splitting the middle term to get the factors for the same. We start solving this problem by finding two numbers such that the product of the two numbers is equal to the product of the coefficient of ${{n}^{2}}$ and the constant. Also, the sum of these two numbers is the coefficient of $n$. Then, after finding these 2 numbers, we split the middle term as the sum of those two numbers and simplify to get the required solution.

Complete step by step solution:
According to the question, we have to find the factors of the quadratic equation.
Thus, we will use splitting the middle term method to get the solution.
The quadratic equation given to us is $42{{n}^{2}}-n-30\to (1)$
As we know, the general form of quadratic equation is $a{{n}^{2}}+bn+c\to (2)$
On comparing equations $(1)$ and $(2)$, we get:
$a=42$
$b=-1$
$c=-30$
To factorize, we have to find two numbers $m$ and $n$ such that $m+n=b$ and $m\times n=a\times c$.
Therefore, the product should be $-1260$ and the sum should be $-1$
We see that if $m=-36$ and $n=35$ , then we get $m+n=-1$ and $m\times n=-1260$
So, we will split the middle term as the addition of $-36n$ and $35n$. On substituting, we get:
$\Rightarrow 42{{n}^{2}}-36n+35n-30$
Now on taking the common terms, we get:
\[\Rightarrow 6n\left( 7n-6 \right)+5\left( 7n-6 \right)\]
Now since the term $\left( 7n-6 \right)$is common in both the terms, we can take it out as common and write the equation as:
\[\Rightarrow \left( 6n+5 \right)\left( 7n-6 \right)\]
Now since there are no more terms which are non-linear, this is the factored form of the equation therefore:
$42{{n}^{2}}-n-30=\left( 6n+5 \right)\left( 7n-6 \right)$, which is the required solution.

Note:
It is not necessary that all the quadratic equations would have roots which are integer numbers or real numbers therefore quadratic formula is used to solve these types of questions, the quadratic formula is:
$({{x}_{1}},{{x}_{2}})=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Where $({{x}_{1}},{{x}_{2}})$are the roots of the equation and $a,b,c$are the coefficients of the quadratic equation.