Solve by Completing the Square x^2-18x+68=0

$x^{2} - 18 x + 68 = 0$

Step 1

Subtract $68$ from both sides of the equation.

$x^{2} - 18 x = - 68$

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 $b$ .

${(\frac{b}{2})}^{2} = {(- 9)}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} - 18 x + {(- 9)}^{2} = - 68 + {(- 9)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Raise $- 9$ to the power of $2$ .

$x^{2} - 18 x + 81 = - 68 + {(- 9)}^{2}$

Simplify the right side.

Simplify $- 68 + {(- 9)}^{2}$ .

Raise $- 9$ to the power of $2$ .

$x^{2} - 18 x + 81 = - 68 + 81$

Add $- 68$ and $81$ .

$x^{2} - 18 x + 81 = 13$

Step 5

Factor the perfect trinomial square into ${(x - 9)}^{2}$ .

${(x - 9)}^{2} = 13$

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.

$x - 9 = \pm \sqrt{13}$

Add $9$ to both sides of the equation.

$x = \pm \sqrt{13} + 9$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm \sqrt{13} + 9$

Decimal Form:

$x = 12.60555127 \dots, 5.39444872 \dots$