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

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

Add $3$ to both sides of the equation.

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

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

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

Add the term to each side of the equation.

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

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 $1$ .

$x^{2} - 2 x + 1 = 4$

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

${(x - 1)}^{2} = 4$

Solve the equation for $x$ .

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

$x - 1 = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

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

Pull terms out from under the radical, assuming positive real numbers.

$x - 1 = \pm 2$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x - 1 = 2$

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

Add $1$ to both sides of the equation.

$x = 2 + 1$

Add $2$ and $1$ .

$x = 3$

Next, use the negative value of the $\pm$ to find the second solution.

$x - 1 = - 2$

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

Add $1$ to both sides of the equation.

$x = - 2 + 1$

Add $- 2$ and $1$ .

$x = - 1$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3, - 1$