College Algebra (6th Edition)

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We substitute 1-2i into x. $(1-2i)^2-2(1-2i)+5$ Take away the parenthesis for the second term. $(1-2i)(1-2i)-2+4i+5$ FOIL it. $1\cdot1-2\cdot{i}+-2i\cdot1+4i\cdot{i}-2+4i+5$ $i\cdot{i}=-1$. $1-2i-2i+4(-1)-2-4i+5$ Group real and imaginary terms. $(1-4-2+5)+(-2-2+4)i$ Add real and imaginary terms. $0+0i$ $0$