How do you solve ${x^2} - 3x + 3 = 0$ by completing the square?

Hint: In this question we have to find the roots from the above quadratic equation by completing the square. For that we are going to simplify the equation. Next, we extract the square root of both sides and then simplify to arrive at our final answer. And also we are going to add and subtract in complete step by step solutions.

Complete step-by-step solution:
To find the roots from the given quadratic equation:
From the given ${x^2} - 3x + 3 = 0$
Divide through by the coefficient of ${x^2}$ if this is not 1. Also, the coefficient of ${x^2}$ is 1 in the given quadratic equation.
Next, move the constant to the right hand side and we get
$ \Rightarrow {x^2} - 3x = - 3$
Then, we add to both sides the square of half the coefficient of $x$:
The square of half the coefficient in this case is${\left( {\dfrac{3}{2}} \right)^2}$.
So we have
$ \Rightarrow {x^2} - 3x + {\left( {\dfrac{3}{2}} \right)^2} = - 3 + {\left( {\dfrac{3}{2}} \right)^2}$
Next, we convert \[{x^2} - 3x + {\left( {\dfrac{3}{2}} \right)^2}\]into the square of a binomial.
That is $ \Rightarrow {x^2} - 3x + {\left( {\dfrac{3}{2}} \right)^2} = {\left( {x - \dfrac{3}{2}} \right)^2}$
Now we have:
$ \Rightarrow {\left( {x - \dfrac{3}{2}} \right)^2} = - 3 + {\left( {\dfrac{3}{2}} \right)^2}$
Next, we squaring the terms on the right hand side (RHS) and we get
 $ \Rightarrow {\left( {x - \dfrac{3}{2}} \right)^2} = - 3 + \left( {\dfrac{9}{4}} \right)$
Now we find the LCM and simplify the right hand side (RHS) and we get
$ \Rightarrow {\left( {x - \dfrac{3}{2}} \right)^2} = \dfrac{{ - 12 + 9}}{4}$
On adding the term and we get
$ \Rightarrow {\left( {x - \dfrac{3}{2}} \right)^2} = \dfrac{{ - 3}}{4}$
Next, we take extract the square root of both sides and we get
$ \Rightarrow \sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2}} = \sqrt {\dfrac{{ - 3}}{4}} $
Already, we know that squares and powers are cancelled. So we apply the rule in the above equation and we get
$ \Rightarrow \left( {x - \dfrac{3}{2}} \right) = \sqrt {\dfrac{{ - 3}}{4}} $
Then, factoring inside the radical on right hand side (RHS) and we get
$ \Rightarrow \sqrt {\dfrac{{ - 3}}{4}} \Rightarrow \sqrt {\left( {\dfrac{1}{4} \times 3 \times \left( { - 1} \right)} \right)} \Rightarrow \sqrt {\dfrac{1}{4}} \times \sqrt 3 \times \sqrt { - 1} $
Here, \[\sqrt {\dfrac{1}{4}} = \dfrac{1}{2}\] and \[\sqrt { - 1} = i\].
Where $i$ is the imaginary unit.
Finally we get
$ \Rightarrow x - \dfrac{3}{2} = \pm \left( {\dfrac{1}{2}\sqrt 3 i} \right)$
Next, we rearrange the variable and numbers on both sides and we get
$ \Rightarrow x = \dfrac{3}{2} \pm \left( {\dfrac{1}{2}\sqrt 3 i} \right)$
Finally we split the above $x$ value we get the required roots.

The roots are
$ \Rightarrow {x_1} = \dfrac{3}{2} + \left( {\dfrac{1}{2}\sqrt 3 i} \right)$
$ \Rightarrow {x_2} = \dfrac{3}{2} - \left( {\dfrac{1}{2}\sqrt 3 i} \right)$

Note: The square of an expression of the form $x + n$ or $x - n$ is :
\[{(x + n)^2} = {x^2} + 2nx + {n^2}\] or \[{(x - n)^2} = {x^2} - 2nx + {n^2}\].
Notice the sign on the middle term matches the sign in the middle of the binomial on the left and the last term is positive in both.
Also notice that if we allow $n$ to be negative, we only need to write and think about\[{(x + n)^2} = {x^2} + 2nx + {n^2}\] (The sign in the middle will match the sign of $n$).
In relation to quadratic equations, imaginary numbers (and complex numbers) occur when the value under the radical portion of the quadratic formula is negative.
When this occurs, the equation has no roots (zeros) in the set of real numbers. The roots belong to the set of complex numbers, and will be called "complex roots" (or "imaginary roots"). These complex roots will be expressed in the form $a + ib$.
The students only remember completing the square. That is a method used to solve a quadratic equation by changing the form of the equation so that the left side is a perfect square trinomial.