What are the solutions to the equation y^2+4y+8=0 ?

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

Find the zeros of y^2+4y+8=0

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

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y^{2} + 4 y + 8 = 0

y = - 2 + 2 i, y = - 2 - 2 i

Solution steps

y^{2} + 4 y + 8 = 0

Solve with the quadratic formula

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

Simplify

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

y_{1, 2} = \frac{- 4 \pm 4 i}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 4 + 4 i}{2 \cdot 1}, y_{2} = \frac{- 4 - 4 i}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

y = - 2 + 2 i, y = - 2 - 2 i

4 x^{2} - 2 x - 7 = 0

x^{2} - 23 x + 22

0 = 0.012 x^{2} - 0.593 x - 14.275

- 5 (v + 4) + 36 \leq 5 (3 - v)

x^{2} + 2.5 x + 60 = 0

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