x² – (2 + i)x – (1 – 7i) = 0

Given x² – (2 + i)x – (1 – 7i) = 0

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

Here, a = 1, b = –(2 + i) and c = –(1 – 7i)

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

We can write 7 – 24i = 16 – 9 – 24i

⇒ 7 – 24i = 16 + 9(–1) – 24i

⇒ 7 – 24i = 16 + 9i² – 24i [∵ i² = –1]

⇒ 7 – 24i = 4² + (3i)² – 2(4)(3i)

⇒ 7 – 24i = (4 – 3i)² [∵ (a – b)² = a² – b² + 2ab]

By using the result 7 – 24i = (4 – 3i)², we get

∴ x = 3 – i or –1 + 2i

Thus, the roots of the given equation are 3 – i and –1 + 2i.