Solve by Completing the Square x^2+10x+27=0

$x^{2} + 10 x + 27 = 0$

Step 1

Subtract $27$ from both sides of the equation.

$x^{2} + 10 x = - 27$

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 $b$ .

${(\frac{b}{2})}^{2} = {(5)}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} + 10 x + {(5)}^{2} = - 27 + {(5)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Raise $5$ to the power of $2$ .

$x^{2} + 10 x + 25 = - 27 + {(5)}^{2}$

Simplify the right side.

Simplify $- 27 + {(5)}^{2}$ .

Raise $5$ to the power of $2$ .

$x^{2} + 10 x + 25 = - 27 + 25$

Add $- 27$ and $25$ .

$x^{2} + 10 x + 25 = - 2$

Step 5

Factor the perfect trinomial square into ${(x + 5)}^{2}$ .

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

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.

$x + 5 = \pm \sqrt{- 2}$

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

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

$x + 5 = \pm \sqrt{- 1 (2)}$

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

$x + 5 = \pm \sqrt{- 1} \cdot \sqrt{2}$

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

$x + 5 = \pm i \sqrt{2}$

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

Subtract $5$ from both sides of the equation.

$x = \pm i \sqrt{2} - 5$

Reorder $\pm i \sqrt{2}$ and $- 5$ .

$x = - 5 \pm i \sqrt{2}$