Solve by Completing the Square x^2-2x+3=0

$x^{2} - 2 x + 3 = 0$

Step 1

Subtract $3$ from both sides of the equation.

$x^{2} - 2 x = - 3$

Step 2

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(- 1)}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} - 2 x + {(- 1)}^{2} = - 3 + {(- 1)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Raise $- 1$ to the power of $2$ .

$x^{2} - 2 x + 1 = - 3 + {(- 1)}^{2}$

Simplify the right side.

Simplify $- 3 + {(- 1)}^{2}$ .

Raise $- 1$ to the power of $2$ .

$x^{2} - 2 x + 1 = - 3 + 1$

Add $- 3$ and $1$ .

$x^{2} - 2 x + 1 = - 2$

Step 5

Factor the perfect trinomial square into ${(x - 1)}^{2}$ .

${(x - 1)}^{2} = - 2$

Step 6

Solve the equation for $x$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x - 1 = \pm \sqrt{- 2}$

Simplify $\pm \sqrt{- 2}$ .

Rewrite $- 2$ as $- 1 (2)$ .

$x - 1 = \pm \sqrt{- 1 (2)}$

Rewrite $\sqrt{- 1 (2)}$ as $\sqrt{- 1} \cdot \sqrt{2}$ .

$x - 1 = \pm \sqrt{- 1} \cdot \sqrt{2}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x - 1 = \pm i \sqrt{2}$

Move all terms not containing $x$ to the right side of the equation.

Add $1$ to both sides of the equation.

$x = \pm i \sqrt{2} + 1$

Reorder $\pm i \sqrt{2}$ and $1$ .

$x = 1 \pm i \sqrt{2}$