Solve by Completing the Square y^2+6y+13=0

$y^{2} + 6 y + 13 = 0$

Step 1

Subtract $13$ from both sides of the equation.

$y^{2} + 6 y = - 13$

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.

$y^{2} + 6 y + {(3)}^{2} = - 13 + {(3)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

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

$y^{2} + 6 y + 9 = - 13 + {(3)}^{2}$

Simplify the right side.

Simplify $- 13 + {(3)}^{2}$ .

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

$y^{2} + 6 y + 9 = - 13 + 9$

Add $- 13$ and $9$ .

$y^{2} + 6 y + 9 = - 4$

Step 5

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

${(y + 3)}^{2} = - 4$

Step 6

Solve the equation for $y$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y + 3 = \pm \sqrt{- 4}$

Simplify $\pm \sqrt{- 4}$ .

Rewrite $- 4$ as $- 1 (4)$ .

$y + 3 = \pm \sqrt{- 1 (4)}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$y + 3 = \pm \sqrt{- 1} \cdot \sqrt{4}$

Rewrite $\sqrt{- 1}$ as $i$ .

$y + 3 = \pm i \cdot \sqrt{4}$

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

$y + 3 = \pm i \cdot \sqrt{2^{2}}$

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

$y + 3 = \pm i \cdot 2$

Move $2$ to the left of $i$ .

$y + 3 = \pm 2 i$

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

Subtract $3$ from both sides of the equation.

$y = \pm 2 i - 3$

Reorder $\pm 2 i$ and $- 3$ .

$y = - 3 \pm 2 i$

Solve by Completing the Square y^2+6y+13=0

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