Solve by Completing the Square x^2-10x+28=0

$x^{2} - 10 x + 28 = 0$

Step 1

Subtract $28$ from both sides of the equation.

$x^{2} - 10 x = - 28$

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.

$x^{2} - 10 x + {(- 5)}^{2} = - 28 + {(- 5)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

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

$x^{2} - 10 x + 25 = - 28 + {(- 5)}^{2}$

Simplify the right side.

Simplify $- 28 + {(- 5)}^{2}$ .

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

$x^{2} - 10 x + 25 = - 28 + 25$

Add $- 28$ and $25$ .

$x^{2} - 10 x + 25 = - 3$

Step 5

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

${(x - 5)}^{2} = - 3$

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.

$x - 5 = \pm \sqrt{- 3}$

Simplify $\pm \sqrt{- 3}$ .

Rewrite $- 3$ as $- 1 (3)$ .

$x - 5 = \pm \sqrt{- 1 (3)}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x - 5 = \pm \sqrt{- 1} \cdot \sqrt{3}$

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

$x - 5 = \pm i \sqrt{3}$

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

Add $5$ to both sides of the equation.

$x = \pm i \sqrt{3} + 5$

Reorder $\pm i \sqrt{3}$ and $5$ .

$x = 5 \pm i \sqrt{3}$