Given x² – 4x + 7 = 0

⇒ x² – 4x + 4 + 3 = 0

⇒ x² – 2(x)(2) + 2² + 3 = 0

⇒ (x – 2)² + 3 = 0 [∵ (a – b)² = a² – 2ab + b²]

⇒ (x – 2)² + 3 × 1 = 0

We have i² = –1 ⇒ 1 = –i²

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

(x – 2)² + 3(–i²) = 0

⇒ (x – 2)² – 3i² = 0

Since a² – b² = (a + b)(a – b), we get

Thus, the roots of the given equation are.