How do you solve ${x^2} - 2x - 12 = 0$ algebraically?

Hint: We will, first of all, write the general quadratic equation and the formula for its roots and then on comparing with the given equation, we have our required answer.

Complete step by step solution:
We are given that we are required to solve ${x^2} - 2x - 12 = 0$ algebraically.
We know that the general quadratic equation is given by $a{x^2} + bx + c = 0$, where a, b and c are constants.
Now, we know that its roots are given by the following expression:-
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Comparing the general equation $a{x^2} + bx + c = 0$ with the given equation ${x^2} - 2x - 12 = 0$, we will then obtain the following:-
$ \Rightarrow $a = 1, b = - 2 and c = - 12
Now, putting these in the formula given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, we will then obtain the following expression:-
$ \Rightarrow x = \dfrac{{ - ( - 2) \pm \sqrt {{{( - 2)}^2} - 4(1)( - 12)} }}{{2(1)}}$
Simplifying the calculations in the right hand side of the above mentioned expression, we will then obtain the following expression:-
$ \Rightarrow x = \dfrac{{2 \pm \sqrt {4 + 48} }}{2}$
Simplifying the calculations in the right hand side of the above mentioned expression further, we will then obtain the following expression:-
$ \Rightarrow x = \dfrac{{2 \pm 2\sqrt {13} }}{2}$
Crossing off 2 from both the numerator and denominator on the right hand side of the above expression, we will then obtain the following expression:-
$ \Rightarrow x = 1 \pm \sqrt {13} $

Note:
The students must note that there is an alternate way to solve the same question.
Alternate way:
We will use the method of completing the square.
We are given that we are required to solve ${x^2} - 2x - 12 = 0$ algebraically.
We can write this as:-
$ \Rightarrow {\left( x \right)^2} - 2 \times 1 \times x + {1^2} - {1^2} - 12 = 0$
Simplifying the calculations a bit, we can write the above expression as:-
$ \Rightarrow {\left( x \right)^2} - 2x + 1 - 13 = 0$ ………………..(1)
Now, we know that we have an identity given by ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$.
Putting x instead of a and 1 instead of b, we will then obtain:-
$ \Rightarrow {\left( {x - 1} \right)^2} = {x^2} + 1 - 2x$
Putting this in equation number 1, we will then obtain:-
$ \Rightarrow {\left( {x - 1} \right)^2} = 13$
Simplifying by taking the square roots on both the sides and re – arranging the terms, we get:-
$ \Rightarrow x = 1 \pm \sqrt {13} $
Thus, we have the roots.