ix² – x + 12i = 0

Given ix² – x + 12i = 0

⇒ ix² + x(–1) + 12i = 0

We have i² = –1

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

ix² + xi² + 12i = 0

⇒ i(x² + ix + 12) = 0

⇒ x² + ix + 12 = 0

⇒ x² + ix – 12(–1) = 0

⇒ x² + ix – 12i² = 0 [∵ i² = –1]

⇒ x² – 3ix + 4ix – 12i² = 0

⇒ x(x – 3i) + 4i(x – 3i) = 0

⇒ (x – 3i)(x + 4i) = 0

⇒ x – 3i = 0 or x + 4i = 0

∴ x = 3i or –4i

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