Solve by Completing the Square k^2+8k+12=0

$k^{2} + 8 k + 12 = 0$

Subtract $12$ from both sides of the equation.

$k^{2} + 8 k = - 12$

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(4)}^{2}$

Add the term to each side of the equation.

$k^{2} + 8 k + {(4)}^{2} = - 12 + {(4)}^{2}$

Simplify the equation.

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Raise $4$ to the power of $2$ .

$k^{2} + 8 k + 16 = - 12 + {(4)}^{2}$

Simplify $- 12 + {(4)}^{2}$ .

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Raise $4$ to the power of $2$ .

$k^{2} + 8 k + 16 = - 12 + 16$

Add $- 12$ and $16$ .

$k^{2} + 8 k + 16 = 4$

Factor the perfect trinomial square into ${(k + 4)}^{2}$ .

${(k + 4)}^{2} = 4$

Solve the equation for $k$ .

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Take the square root of each side of the equation to set up the solution for $k$

${(k + 4)}^{2 \cdot \frac{1}{2}} = \pm \sqrt{4}$

Remove the perfect root factor $k + 4$ under the radical to solve for $k$ .

$k + 4 = \pm \sqrt{4}$

Simplify the right side of the equation.

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

$k + 4 = \pm \sqrt{2^{2}}$

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

$k + 4 = \pm 2$

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

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

$k + 4 = 2$

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

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Subtract $4$ from both sides of the equation.

$k = 2 - 4$

Subtract $4$ from $2$ .

$k = - 2$

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

$k + 4 = - 2$

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

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Subtract $4$ from both sides of the equation.

$k = - 2 - 4$

Subtract $4$ from $- 2$ .

$k = - 6$

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

$k = - 2, - 6$

Solve by Completing the Square k^2+8k+12=0