What are the solutions to the equation y^2+8y+41=0 ?

The solutions to the equation y^2+8y+41=0 are y=-4+5i,y=-4-5i

Find the zeros of y^2+8y+41=0

The zeros of y^2+8y+41=0 are y=-4+5i,y=-4-5i

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

y = - 4 + 5 i, y = - 4 - 5 i

Solution steps

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

Solve with the quadratic formula

y_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 41}{2 \cdot 1}

Simplify

8^{2} - 4 \cdot 1 \cdot 41 : 10 i

y_{1, 2} = \frac{- 8 \pm 10 i}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 8 + 10 i}{2 \cdot 1}, y_{2} = \frac{- 8 - 10 i}{2 \cdot 1}

y = \frac{- 8 + 10 i}{2 \cdot 1} : - 4 + 5 i

y = \frac{- 8 - 10 i}{2 \cdot 1} : - 4 - 5 i

The solutions to the quadratic equation are:

y = - 4 + 5 i, y = - 4 - 5 i

\frac{1}{2 !}

y, x^{3} - x y + y^{2} = 7

x^{2} \leq 8 x

1 1^{4}

\frac{1}{3 n} = 4

What are the solutions to the equation y^2+8y+41=0 ?
The solutions to the equation y^2+8y+41=0 are y=-4+5i,y=-4-5i
Find the zeros of y^2+8y+41=0
The zeros of y^2+8y+41=0 are y=-4+5i,y=-4-5i