#### Answer

$0$

#### Work Step by Step

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.
$0+0i$
$0$