Question 305085
x^2 + 6x + 23 = 0

Rewrie the above as:

x^2 + 6x = -23

To complete the square on the loeft side we need to divide the coefficient on the x term by 2,  square it and add the result to both sides:

The coefficient on the x term is 6, so we have (6/2)^2 = 3^2 = 9. Adding 9 to both sides we have:

x^2 + 6x + 9 = -23 + 9

x^2 + 6x + 9 = -14

(x+3)*(x+3) = -14
(x+3)^2 = -14

Taking the square root of both sides:

(x+3)= +sqrt(-14) and
(x+3)= -sqrt(-14)

Note that sqrt(-14) is a complex number, namely i*sqrt(14).

We have then:

x+3 = i*sqrt(14)
x = -3 + i*sqrt914)

and 

x+3 = -i*sqrt(14)
x = -3 - i*sqrt(14)