Solve by Completing the Square y^2+8y+41=0

$y^{2} + 8 y + 41 = 0$

Subtract $41$ from both sides of the equation.

$y^{2} + 8 y = - 41$

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.

$y^{2} + 8 y + {(4)}^{2} = - 41 + {(4)}^{2}$

Simplify the equation.

Simplify the left side.

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

$y^{2} + 8 y + 16 = - 41 + {(4)}^{2}$

Simplify the right side.

Simplify $- 41 + {(4)}^{2}$ .

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

$y^{2} + 8 y + 16 = - 41 + 16$

Add $- 41$ and $16$ .

$y^{2} + 8 y + 16 = - 25$

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

${(y + 4)}^{2} = - 25$

Solve the equation for $y$ .

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

$y + 4 = \pm \sqrt{- 25}$

Simplify $\pm \sqrt{- 25}$ .

Rewrite $- 25$ as $- 1 (25)$ .

$y + 4 = \pm \sqrt{- 1 (25)}$

Rewrite $\sqrt{- 1 (25)}$ as $\sqrt{- 1} \cdot \sqrt{25}$ .

$y + 4 = \pm \sqrt{- 1} \cdot \sqrt{25}$

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

$y + 4 = \pm i \cdot \sqrt{25}$

Rewrite $25$ as $5^{2}$ .

$y + 4 = \pm i \cdot \sqrt{5^{2}}$

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

$y + 4 = \pm i \cdot 5$

Move $5$ to the left of $i$ .

$y + 4 = \pm 5 i$

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

Subtract $4$ from both sides of the equation.

$y = \pm 5 i - 4$

Reorder $\pm 5 i$ and $- 4$ .

$y = - 4 \pm 5 i$