Question 1007384
given:
{{{x[1]=2i}}}
then we also have {{{x[2]=-2i}}} (complex roots  always come in pairs)

{{{(x-x[1])(x-x[2])=0}}}

{{{(x-2i)(x-(-2i))=0}}}

{{{(x-2i)(x+2i)=0}}}

{{{x^2-(2i)^2=0}}}

{{{x^2-4i^2=0}}}

{{{x^2-4(-1)=0}}}

{{{x^2+4=0}}}