How do you use a quadratic formula to solve: $ {x^2} - 2x + 2 = 0 $ ?

Hint: The given problem requires us to solve a quadratic equation using quadratic formula. There are various methods that can be employed to solve a quadratic equation like completing the square method, using quadratic formulas and by splitting the middle term. Using the quadratic formula gives us the roots of the equation directly with ease.

Complete step-by-step answer:
In the given question, we are required to solve the equation $ {x^2} - 2x + 2 = 0 $ with the help of quadratic formula. The quadratic formula can be employed for solving an equation only if we compare the given equation with the standard form of quadratic equation.
Comparing with standard quadratic equation $ a{x^2} + bx + c = 0 $
Here, $ a = 1 $ , $ b = - 2 $ and $ c = 2 $ .
Now, Using the quadratic formula, we get the roots of the equation as:
 $ x = \dfrac{{( - b) \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Substituting the values of a, b, and c in the quadratic formula, we get,
 $ \Rightarrow x = \dfrac{{\left( { - \left( { - 2} \right)} \right) \pm \sqrt {{{( - 2)}^2} - 4 \times 1 \times \left( 2 \right)} }}{{2 \times 1}} $
 $ \Rightarrow x = \dfrac{{\left( 2 \right) \pm \sqrt {4 - 8} }}{2} $
 $ \Rightarrow x = \dfrac{{2 \pm \sqrt { - 4} }}{2} $
The discriminant of the given equation is negative. Hence, the roots of the equation are not real. The roots belong to the set of complex numbers. So, we get,
 $ \Rightarrow x = \dfrac{{2 \pm 2i}}{2} $
 $ \Rightarrow x = 1 \pm i $
So, $ x = 1 + i $ and $ x = 1 - i $ are the roots of the equation $ {x^2} - 2x + 2 = 0 $ .
So, the correct answer is “ $ x = 1 + i $ and $ x = 1 - i $ ”.

Note: Quadratic equations are the polynomial equations with degree of the variable or unknown as 2. Quadratic formula is the easiest and most efficient formula to calculate the roots of an equation. Quadratic equations can also be solved by splitting the middle term and completing the square method. Quadratic equations may also be solved by hit and trial method if the roots of the equation are easy to find.