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**Hint:**Before solving the given equation, we have to regroup it. We are given the equation in the form $a{x^2} + bx + c = 0$, here, we have the value of ‘$b$’ and we have to divide the value of ‘$b$’ by $2$. We then have to square the obtained value. Next, we add and subtract the value from the given equation. We even use the formula ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$. By using these steps, we can solve the given problem.

**Complete step by step solution:**

Given equation, ${x^2} - 4x + 5 = 0$.First, we regroup the given equation,

$\left( {{x^2} - 4x} \right) + 5 = 0$----(1)

The given equation is in the standard form $a{x^2} + bx + c = 0$, where $a = 1$ and $b = - 4$.Next, we now divide ‘$b$’ by $2$,

$\dfrac{{ - 4}}{2} = - 2$

Now we have to square the obtained value i.e., ${\left( { - 2} \right)^2} = 4$. We now add and subtract the obtained value from equation (1).

$

\left( {{x^2} - 4x + 4} \right) + 5 - 4 = 0 \\

\Rightarrow \left( {{x^2} - 4x + 4} \right) + 1 = 0 \\ $

We can clearly see that ${x^2} - 4x + 4$ is in the form of ${a^2} - 2ab + {b^2}$ with $a = x$ and $b = - 2$. So, we can rewrite the obtained equation into the form ${\left( {a + b} \right)^2}$. That is ${\left( {x - 2} \right)^2}$.

$ \Rightarrow {\left( {x - 2} \right)^2} + 1 = 0$

Now, subtracting $1$ from both sides of the equation and we get,

$

\Rightarrow {\left( {x - 2} \right)^2} + 1 - 1 = 0 - 1 \\

\Rightarrow {\left( {x - 2} \right)^2} = - 1 \\ $

Now, after taking square root on both the sides of the equation, we get,

$ \Rightarrow \sqrt {{{\left( {x - 2} \right)}^2}} = \sqrt { - 1} $

We know that, square and square root cancels out,

$

\left( {x - 2} \right) = \pm \sqrt { - 1} \\

\therefore \left( {x - 2} \right) = \pm \iota \\ $

Iota ($\iota $) is not a real number but is an imaginary number.

**Hence, there are no real roots of the given equation.**

**Note:**The same procedure is followed while solving similar problems. Students need to be extra careful in the calculation part. They should also be careful while adding the obtained value after squaring, we have to balance the equation by subtracting the same number too. This question could also be solved by a method of factorization.

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