Solve by Completing the Square x^2-6x-40=0

$x^{2} - 6 x - 40 = 0$

Step 1

Add $40$ to both sides of the equation.

$x^{2} - 6 x = 40$

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.

$x^{2} - 6 x + {(- 3)}^{2} = 40 + {(- 3)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

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

$x^{2} - 6 x + 9 = 40 + {(- 3)}^{2}$

Simplify the right side.

Simplify $40 + {(- 3)}^{2}$ .

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

$x^{2} - 6 x + 9 = 40 + 9$

Add $40$ and $9$ .

$x^{2} - 6 x + 9 = 49$

Step 5

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

${(x - 3)}^{2} = 49$

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.

$x - 3 = \pm \sqrt{49}$

Simplify $\pm \sqrt{49}$ .

Rewrite $49$ as $7^{2}$ .

$x - 3 = \pm \sqrt{7^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x - 3 = \pm 7$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x - 3 = 7$

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

Add $3$ to both sides of the equation.

$x = 7 + 3$

Add $7$ and $3$ .

$x = 10$

Next, use the negative value of the $\pm$ to find the second solution.

$x - 3 = - 7$

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

Add $3$ to both sides of the equation.

$x = - 7 + 3$

Add $- 7$ and $3$ .

$x = - 4$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 10, - 4$

Solve by Completing the Square x^2-6x-40=0

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