Question 362320
X=2+i square root of 11
{{{x-2=i*sqrt(11)}}}
square 
(x-2)^2=i^2*11
x^2-4x+4=-1*11
x^2-4x+4 =-11
x^2-4x+15=0
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m.ananth@hotmail.ca
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X= 2-i square root of 11
x-2=-i*sqrt11
(x-2)^2=-i^2*11
x^2-4x+4=11
x^2-4x-7=0