## College Algebra (10th Edition)

The solution set is $\color{blue}{\left\{1-2i, 1+2i\right\}}$.
Subtract $5$ to both sides of the equation: $x^2-2x+5-5=0-5 \\x^2-2x=-5$ Complete the square by adding $(\frac{-2}{2})^2=1$ to both sides of the equation: $x^2-2x+1 = -5+1 \\x^2-2x+1=-4$ Write the trinomial in factored form to obtain: $(x-1)^2=-4$ Take the square root of both sides: $\sqrt{(x-1)^2} = \pm \sqrt{-4} \\x-1=\pm \sqrt{-4} \\x-1=\pm \sqrt{4(-1)} \\x-1=\pm \sqrt{2^2(-1)} \\x-1=\pm 2\sqrt{-1}$ Add $1$ to both sides of the equation to obtain: $x-1+1=1\pm2\sqrt{-1} \\x=1\pm 2\sqrt{-1}$ Note that $\sqrt{-1} = i$. Thus, the equation above simplifies to: $x=1 \pm 2i$ Therefore, the solution set is $\color{blue}{\left\{1-2i, 1+2i\right\}}$.