Solve by Completing the Square x^2+2x-5=0

$x^{2} + 2 x - 5 = 0$

Step 1

Add $5$ to both sides of the equation.

$x^{2} + 2 x = 5$

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.

$x^{2} + 2 x + {(1)}^{2} = 5 + {(1)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

One to any power is one.

$x^{2} + 2 x + 1 = 5 + {(1)}^{2}$

Simplify the right side.

Simplify $5 + {(1)}^{2}$ .

One to any power is one.

$x^{2} + 2 x + 1 = 5 + 1$

Add $5$ and $1$ .

$x^{2} + 2 x + 1 = 6$

Step 5

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

${(x + 1)}^{2} = 6$

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.

$x + 1 = \pm \sqrt{6}$

Subtract $1$ from both sides of the equation.

$x = \pm \sqrt{6} - 1$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm \sqrt{6} - 1$

Decimal Form:

$x = 1.44948974 \dots, - 3.44948974 \dots$