We have to find the roots of 9x^2 - 6x + 21 = 0 by completing the square.

9x^2 - 6x + 21 = 0

=> (3x)^2 - 2*3x + 1 = -20

=> (3x - 1)^2 = -20

=> 3x - 1= sqrt -20 and 3x + 1 = -sqrt (-20)

=> 3x = 1 + i*sqrt 20 and 3x = 1 - i*sqrt 20

=> x = 1/3 + i*(sqrt 20)/3 and x = 1/3 - i*(sqrt 20)/3

**The required roots are x = 1/3 + i*(sqrt 20)/3 and x = 1/3 - i*(sqrt 20)/3**

We'll start by subtracting 21 both sides, to move the number alone to the right side of the equation, so that being more clear what we have to add to the left side, to complete the square.

9x^2 - 6x = -21

We'll complete the square by adding the number 1 to both side, to get a perfect square to the left side.

9x^2 -6x +1 = -21 + 1

We'll write the left side as a perfect square:

(3x - 1)^2 = -20

3x - 1 = sqrt -20

3x - 1 = 2isqrt5

x1 = (1 + 2isqrt5)/3

x1 = 1/3 + (2sqrt5/3)*i

x2 = 1/3 - (2sqrt5/3)*i

**The complex solutions of the equation are: { 1/3 + (2sqrt5/3)*i ; 1/3 - (2sqrt5/3)*i}.**