How do you solve $\left( {x - 5} \right)\left( {x + 6} \right) = 12$?

Hint: First, expand the terms on the left side. After that move the constant part on the left side. Then the equation is quadratic in one variable, we will use the quadratic formula to find the roots of the given equation. If the given quadratic equation is of the form $a{x^2} + bx + c = 0$, the quadratic formula is given as $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

Complete step-by-step answer:
We have been given an equation $\left( {x - 5} \right)\left( {x + 6} \right) = 12$.
We have to find the roots of the given equation.
First, expand the terms on the left side,
$ \Rightarrow {x^2} + x - 30 = 12$
Now, move the constant part on the left side,
$ \Rightarrow {x^2} + x - 30 - 12 = 0$
Simplify the terms,
$ \Rightarrow {x^2} + x - 42 = 0$
We will compare the given equation with the standard quadratic equation which is given by $a{x^2} + bx + c = 0$.
On comparing we get the values $a = 1,b = 1,c = - 42$.
Now, we know that the quadratic formula is given as
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substituting the values in the above formula we get
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {{{\left( 1 \right)}^2} - 4 \times 1 \times - 42} }}{{2 \times 1}}$
Now, on solving the obtained equation we get
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 + 168} }}{2}$
Add the terms in the square root,
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {169} }}{2}$
Now, we know that the value of the square root $\sqrt {169} = 13$.
$ \Rightarrow x = \dfrac{{ - 1 \pm 13}}{2}$
Now, we know that a quadratic equation has two roots. We can write the obtained equation as
$ \Rightarrow x = \dfrac{{ - 1 + 13}}{2}$ and $x = \dfrac{{ - 1 - 13}}{2}$
Simplify the terms,
$ \Rightarrow x = \dfrac{{12}}{2}$ and $x = \dfrac{{ - 14}}{2}$
Cancel out the common factors,
$ \Rightarrow x = 6$ and $x = - 7$

Hence, the two roots of the equation $\left( {x - 5} \right)\left( {x + 6} \right) = 12$ is -7, and 6.

Note:
Avoid calculation mistakes because single calculation mistakes lead to an incorrect answer. To solve a quadratic equation, students can use the factorization method, completing the square method or quadratic formula method. When the time is less and we are sure about the quadratic formula, then it is best to use this method. We can cross verify the factors by opening parenthesis and solving.