Solve ${\left( {x - 8} \right)^2} = 144$?

Hint: In this question we have to solve the given polynomial using quadratic formula. First we will expand the square by using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$. Then we will get an quadratic polynomial in the form $a{x^2} + bx + c$, where "$a$", "$b$", and “$c$" are real numbers and the Quadratic Formula is derived from the process of completing the square, and is formally stated as:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

Complete step by step answer:
Given quadratic equation is,
${\left( {x - 8} \right)^2} = 144$,
Using the identity and rewrite the left side of the equation, we get,
$ \Rightarrow {\left( {x - 8} \right)^2} = {\left( x \right)^2} - 2\left( x \right)\left( 8 \right) + {\left( 8 \right)^2} = 144$,
$ \Rightarrow {x^2} - 16x + 64 = 144$,
Subtract 144 from both sides of the equation we get,
$ \Rightarrow {x^2} - 16x + 64-144 = 144 - 144$,
$ \Rightarrow {x^2} - 16x - 80 = 0$,
Use the quadratic formula, which is given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}, a = 1$
Here $a = 1$,$b = - 16$,$c = - 80$,
Now substituting the values in the formula we get,
$ \Rightarrow x = \dfrac{{ - \left( { - 16} \right) \pm \sqrt {{{\left( { - 16} \right)}^2} - 4\left( 1 \right)\left( { - 80} \right)} }}{{2\left( 1 \right)}}$,
On simplifying we get,
$ \Rightarrow x = \dfrac{{16 \pm \sqrt {256 - \left( { - 320} \right)} }}{2}$,
$ \Rightarrow x = \dfrac{{16 \pm \sqrt {256 + 320} }}{2}$,
$ \Rightarrow x = \dfrac{{16 \pm \sqrt {576} }}{2}$,
Taking the square root we get,
$ \Rightarrow x = \dfrac{{16 \pm 24}}{2}$,
Now we get two values if they are $x = \dfrac{{16 + 24}}{2} = \dfrac{{ 40}}{2} = 20$.
And, $x = \dfrac{{16 - 24}}{2} = \dfrac{{ - 8}}{2} = - 4$.
So the values of $x$ are -4 and 20.

If we solve the given equation, i.e., ${\left( {x - 8} \right)^2} = 144$, then the values of $x$ are -4 and 20.

Note: Quadratic equation formula is a method of solving quadratic equations, but we should keep in mind that we can also solve the equation using completely the square, and we can cross check the values of $x$ by using the above formula. Also we should always convert the coefficient of ${x^2} = 1$, to easily solve the equation by this method, and there are other methods to solve such kind of solutions, other method used to solve the quadratic equation is by factoring method, in this method we should obtain the solution factorising quadratic equation terms. In these types of questions, we can solve by using quadratic formula i.e., $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.