How do you complete the square to solve $9{x^2} - 12x + 5 = 0$?

Hint: We will first make the term with square of x as to be square of something and try to make it in the form of \[{(a - b)^2} = {a^2} + {b^2} - 2ab\].

Complete step-by-step answer:
We are given that we are required to solve $9{x^2} - 12x + 5 = 0$ using the method of completing the squares.
First of all, we notice that we are given the square of x with a coefficient of 9 in the given quadratic. And, we know that $9{x^2} = {(3x)^2}$.
So, we can write the given quadratic equation as ${(3x)^2} - 12x + 5 = 0$
Now, we will modify the term with a coefficient of x by making it in the form of multiple of 2 and 3x.
So, we can write the given quadratic equation as ${(3x)^2} - 2 \times 3x \times 2 + 5 = 0$
Now, we will just modify the left out constant term five in terms of square of 2.
So, we can write the given quadratic equation as ${(3x)^2} - 2 \times 3x \times 2 + {(2)^2} + 1 = 0$ …….(1)
Now, since we know that we have a formula given by: \[{(a - b)^2} = {a^2} + {b^2} - 2ab\].
Replacing a by 3x and b by 2, we will then obtain the following equation:-
\[ \Rightarrow {(3x - 2)^2} = {(3x)^2} + {2^2} - 2 \times 2 \times 3x\]
Now, if we look at the equation number 1, we will get:-
The given quadratic equation can be written as ${(3x - 2)^2} - 1 = 0$
Now, taking 1 to right hand side, and taking the square root of both sides, we will then get:-
\[ \Rightarrow 3x - 2 = \pm 1\]
Taking 2 to right hand side, we will then obtain:-
Either \[3x = 1\] or \[3x = 3\]

Either \[x = \dfrac{1}{3}\] or \[x = 1\].

Note:
The students must note that they could have tackled the coefficient of square of x that is 9 in the beginning only by dividing both sides by 9 and then following the same as given above.
If we would not have had given the method of completing square, we could have used an alternate method given as follows:-
$ \Rightarrow 9{x^2} - 12x + 5 = 0$
$ \Rightarrow 9{x^2} + 3x - 15x + 5 = 0$
$ \Rightarrow 3x(3x + 1) - 5(3x + 1) = 0$
$ \Rightarrow (3x + 1)(3x - 5) = 0$
Thus we have the roots.