x² + 10ix – 21 = 0

Given x² + 10ix – 21 = 0

⇒ x² + 10ix – 21 × 1 = 0

We have i² = –1 ⇒ 1 = –i²

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

x² + 10ix – 21(–i²) = 0

⇒ x² + 10ix + 21i² = 0

⇒ x² + 3ix + 7ix + 21i² = 0

⇒ x(x + 3i) + 7i(x + 3i) = 0

⇒ (x + 3i)(x + 7i) = 0

⇒ x + 3i = 0 or x + 7i = 0

∴ x = –3i or –7i

Thus, the roots of the given equation are –3i and –7i.