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

The solutions to the equation y^2+2y+2=0 are y=-1+i,y=-1-i

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

The zeros of y^2+2y+2=0 are y=-1+i,y=-1-i

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

y = - 1 + i, y = - 1 - i

Solution steps

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = - 1 + i, y = - 1 - i

x^{2} - x = 0

2^{n} - 2^{n - 1}

∣ 3 (x - 1) + 2 ∣ \leq 20

ln (x y)

2 (4 u - 7) = 66

What are the solutions to the equation y^2+2y+2=0 ?
The solutions to the equation y^2+2y+2=0 are y=-1+i,y=-1-i
Find the zeros of y^2+2y+2=0
The zeros of y^2+2y+2=0 are y=-1+i,y=-1-i