How do you solve this algebra 2 question

y^2 - 3y + 3 = 0

... y^2 - 3y + 3 = 0
The Quadratic Formula:
x = [ -b ± √ ( (b)² - 4(a)(c) ) ] / 2(a)
x = [ -(-3) ± √ ( (-3)² - 4(1)(3) ) ] / 2(1)
x = [ 3 ± √ ( 9 - 12 ) ] / 2
x = [ 3 ± √3 i ] / 2
x = { ½ (3-√3 i), ½ (3+√3 i) }

use the quadratic formula

(3 ± √-3) / 2

y = (3 ± i√3) / 2

two imaginary solutions

Complete the square.