Answer
$\left\{1-\sqrt3, 1+\sqrt3\right\}$
Work Step by Step
RECALL:
To solve $x^2 + bx=c$ by completing the square, add $\left(\frac{b}{2}\right)^2$ on both sides of the equation.
Thus, to solve the given equation by completing the square, add $(\frac{-2}{2})^2=(-1)^2=1$ on both sides of the equation to obtain:
$x^2-2x+1=2+1
\\x^2-2x+1=3
\\(x-1)^2=3$
Take the square root of both sides to obtain:
$x-1=\pm \sqrt{3}$
Add 1 on both sides of the equation to obtain:
$x = 1 \pm\sqrt3$
Two solutions:
$x_1 = 1+\sqrt3
\\x_2 = 1-\sqrt3$
Therefore, the solution of the given equation is $\left\{1-\sqrt3, 1+\sqrt3\right\}$.
