Given 4x² – 12x + 25 = 0

⇒ 4x² – 12x + 9 + 16 = 0

⇒ (2x)² – 2(2x)(3) + 3² + 16 = 0

⇒ (2x – 3)² + 16 = 0 [∵ (a + b)² = a² + 2ab + b²]

⇒ (2x – 3)² + 16 × 1 = 0

We have i² = –1 ⇒ 1 = –i²

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

(2x – 3)² + 16(–i²) = 0

⇒ (2x – 3)² – 16i² = 0

⇒ (2x – 3)² – (4i)² = 0

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

(2x – 3 + 4i)(2x – 3 – 4i) = 0

⇒ 2x – 3 + 4i = 0 or 2x – 3 – 4i = 0

⇒ 2x = 3 – 4i or 2x = 3 + 4i

Thus, the roots of the given equation are.