Given x² + 2x + 2 = 0

⇒ x² + 2x + 1 + 1 = 0

⇒ x² + 2(x)(1) + 1² + 1 = 0

⇒ (x + 1)² + 1 = 0 [∵ (a + b)² = a² + 2ab + b²]

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

(x + 1)² + (–i²) = 0

⇒ (x + 1)² – i² = 0

⇒ (x + 1)² – (i)² = 0

⇒ (x + 1 + i)(x + 1 – i) = 0 [∵ a² – b² = (a + b)(a – b)]

⇒ x + 1 + i = 0 or x + 1 – i = 0

⇒ x = –1 – i or x = –1 + i

∴ x = –1 ± i

Thus, the roots of the given equation are –1 ± i.