x² – (5 – i)x + (18 + i) = 0

Given x² – (5 – i)x + (18 + i) = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 1, b = –(5 – i) and c = (18 + i)

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

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

We can write 48 + 14i = 49 – 1 + 14i

⇒ 48 + 14i = 49 + i² + 14i [∵ i² = –1]

⇒ 48 + 14i = 7² + i² + 2(7)(i)

⇒ 48 + 14i = (7 + i)² [∵ (a + b)² = a² + b² + 2ab]

By using the result 48 + 14i = (7 + i)², we get

[∵ i² = –1]

∴ x = 2 + 3i or 3 – 4i

Thus, the roots of the given equation are 2 + 3i and 3 – 4i.