Chapter 2 - Linear and Quadratic Functions - Section 2.8 Equations and Inequalities Involving the Absolute Value Function - 2.8 Assess Your Understanding - Page 182: 35

$-1, 3, 1\pm i\sqrt 2$

Step 1. The original equation can be separated into two cases: $x^2-2x=3$ or $x^2-2x=-3$. Step 2. For $x^2-2x=3$, we have $x^2-2x-3=0 \longrightarrow (x+1)(x-3)=0 \longrightarrow x=-1, 3$. Step 3. For $x^2-2x=-3$, we have $x^2-2x+3=0 \longrightarrow x=\frac{2\pm\sqrt {4-4(3)}}{2}=1\pm i\sqrt 2$.