Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 593: 42

{$2-2i,2+2i$}

Given: $x^2-4x+8=0$ This can be re-written as: $x^2-4x=-8$ We will have to add both sides $(-2)^2$ to complete the square. Thus, $x^2-4x+(-2)^2=-8+(-2)^2$ or, $x^2-4x+4=-4$ or, $(x-2)^2=-4$ or, $x-2=\pm \sqrt {-4} \implies x =2 \pm 2i$ Hence, our desired solution set is {$2-2i,2+2i$}