How do you solve using the quadratic formula ${{x}^{2}}+x-42=0$?

Hint: We are given ${{x}^{2}}+x-42=0$ to solve this we learn about the type of Equation we are given then learn the number of solutions of the equation. We will learn how to factor the quadratic equation, we will use the middle term split to factor the term and we will simplify by taking common terms out. We also use zero product rules to get our answer. To be sure about your answer we can also check by putting the acquired value of the solution in the given Equation and check whether they are the same or not.

Complete step-by-step solution:
We are asked to solve the given problem ${{x}^{2}}+x-42=0$. First, we observe that it has a maximum power of $2$ so it is a quadratic equation.
Now we should know that a quadratic equation has a $2$ solution or we say an equation of power 'n' will have an 'n' solution.
Now as it is a quadratic equation, we will change it into standard form \[\mathbf{a}{{\mathbf{y}}^{2}}+\mathbf{by}+\mathbf{c}=\mathbf{0}\]\[\]
As we look closely our problem is already in standard form ${{x}^{2}}+x-42=0$
Now we have to solve the equation ${{x}^{2}}+x-42=0$
To solve this equation, we first take the greatest common factor possibly available to the terms.
As we can see that in ${{x}^{2}}+x-42=0$
\[1,1,-42\] has nothing in common
Equation remains same
$\Rightarrow {{x}^{2}}+x-42=0$
Now, as we are asked, we have to solve using the Quadratic formula so we should know what quadratic formula mean for ay Quadratic equation \[\mathbf{a}{{\mathbf{y}}^{2}}+\mathbf{b}x+\mathbf{c}=\mathbf{0}\]
Solution is given by the formula
$\begin{align}
  & x=\dfrac{-b\pm \sqrt{4ac}}{2a} \\
 & \\
\end{align}$
This formula is called quadratic formula for our equation
${{x}^{2}}+x-42=0$
We have $a=1,\,b=1,\,c=-42$
So using this in quadratic formula we get,
$x=\dfrac{-1\pm \sqrt{{{(1)}^{2}}-4\times 1\times 1-42}}{2(1)}$
Simplifying we get
$\Rightarrow x=\dfrac{-1\pm \sqrt{169}}{2}$
As $\sqrt{169}=13$
\[\Rightarrow x=\dfrac{-1\pm 13}{2}\]
Hence, we get solution as
\[x=\dfrac{-1+13}{2},\,\,\,x=\dfrac{-1\pm -3}{2}\]
So, solution for ${{x}^{2}}+x-42=0$ are
\[x=6\,\,\,\,\text{and}\,\,\,x=-7\]

Note: This is a good method to solve our pattern based on quadratic equations. It gives answers very directly and it causes less chances of error. We can check our answer by putting \[x=6\,\,\,\,\text{and}\,\,\,x=-7\] in ${{x}^{2}}+x-42=0$. If it satisfies that our solution is correct putting
\[x=6\,\,\operatorname{in}\,\,{{x}^{2}}+x-42=0\]
We get,
\[\begin{align}
  & {{6}^{2}}+6-42=0 \\
 & 42-42=0 \\
\end{align}\]
It is true.
So \[x=6\] is correct.
Now checking \[x=-7\]
Putting \[x=-7\,\operatorname{in}\,\,{{x}^{2}}+x-42=0\]
\[\begin{align}
  & {{(-7)}^{2}}-7-42=0 \\
 & 49-49=0 \\
\end{align}\]
Which is also true.
So, \[x=-7\] is also the right solution.