Solve by Completing the Square x^2-14x+47=0

$x^{2} - 14 x + 47 = 0$

Step 1

Subtract $47$ from both sides of the equation.

$x^{2} - 14 x = - 47$

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 $b$ .

${(\frac{b}{2})}^{2} = {(- 7)}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} - 14 x + {(- 7)}^{2} = - 47 + {(- 7)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Raise $- 7$ to the power of $2$ .

$x^{2} - 14 x + 49 = - 47 + {(- 7)}^{2}$

Simplify the right side.

Simplify $- 47 + {(- 7)}^{2}$ .

Raise $- 7$ to the power of $2$ .

$x^{2} - 14 x + 49 = - 47 + 49$

Add $- 47$ and $49$ .

$x^{2} - 14 x + 49 = 2$

Step 5

Factor the perfect trinomial square into ${(x - 7)}^{2}$ .

${(x - 7)}^{2} = 2$

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.

$x - 7 = \pm \sqrt{2}$

Add $7$ to both sides of the equation.

$x = \pm \sqrt{2} + 7$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm \sqrt{2} + 7$

Decimal Form:

$x = 8.41421356 \dots, 5.58578643 \dots$