Given the equation :

9x^2 -6x +21 = 0

We need to solve by completing the square.

==> Let us rewrite 21 = 21 +1

==> 9x^2 -6x + 1 + 20 = 0

==> (3x-1)^2 + 20 = 0

==> Now we will move 20 to the right side.

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

Now we will take the square root of both sides.

==> 3x-1 = sqrt-20 = 2sqrt-5 = 2sqrt5*i

==> 3x-1 =2sqrt5*i

==> 3x = 1+ 2sqrt5*i

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

OR :

-(3x-1) = 2sqrt5*i

==> -3x +1 = 2sqrt5*i

==> -3x = -1 + 2sqrt5*i

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

Then we have two solutions:

**x= { (1/3)+2sqrt5/3 *i , (1/3) 8 (2sqrt5/3)*i }**

We have the equation 9x^2-6x+21=0 to solve for x by completion of the square.

9x^2 - 6x + 21 = 0

=> (3x)^2 - (3x)*2*1 + 1 + 20 = 0

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

=> 3x - 1 = i*sqrt 20

=> 3x = 1 + i*sqrt 20

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

and

3x - 1 = -i*sqrt 20

=> 3x = 1 - i*sqrt 20

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

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

To complete the square and find x. 9x^2-6x+21=0.

9x^2-6x+21=0.

Subtract 21 from both sides:

9x^2-6x = -21

We add 1 to both sides so that left side becomes (9x^2-6x+1 = 3x-1)^2.

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

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

We take the square root.

3x-1 = +sqrt(-20) = 2sqrt(-5).

Or3x-1 = -sqrt(-20) = -2sqrt(-5).

3x= 2srt(-5)+1.

So **x = {1+2sqrt(-5)}/3.**

Similarly 3x-1 = - sqrt(-5) gives**x = (1-sqrt(-5)}/3.**

For the beginning, we'll subtract 21 both sides, to move the constant on 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}.**