Solve by Completing the Square x^2+6x-59=0

$x^{2} + 6 x - 59 = 0$

Step 1

Add $59$ to both sides of the equation.

$x^{2} + 6 x = 59$

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 $b$ .

${(\frac{b}{2})}^{2} = {(3)}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} + 6 x + {(3)}^{2} = 59 + {(3)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Raise $3$ to the power of $2$ .

$x^{2} + 6 x + 9 = 59 + {(3)}^{2}$

Simplify the right side.

Simplify $59 + {(3)}^{2}$ .

Raise $3$ to the power of $2$ .

$x^{2} + 6 x + 9 = 59 + 9$

Add $59$ and $9$ .

$x^{2} + 6 x + 9 = 68$

Step 5

Factor the perfect trinomial square into ${(x + 3)}^{2}$ .

${(x + 3)}^{2} = 68$

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.

$x + 3 = \pm \sqrt{68}$

Simplify $\pm \sqrt{68}$ .

Rewrite $68$ as $2^{2} \cdot 17$ .

Factor $4$ out of $68$ .

$x + 3 = \pm \sqrt{4 (17)}$

Rewrite $4$ as $2^{2}$ .

$x + 3 = \pm \sqrt{2^{2} \cdot 17}$

Pull terms out from under the radical.

$x + 3 = \pm 2 \sqrt{17}$

Subtract $3$ from both sides of the equation.

$x = \pm 2 \sqrt{17} - 3$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm 2 \sqrt{17} - 3$

Decimal Form:

$x = 5.24621125 \dots, - 11.24621125 \dots$