What are the solutions to the equation y^2+2y+5=0 ?

The solutions to the equation y^2+2y+5=0 are y=-1+2i,y=-1-2i

Find the zeros of y^2+2y+5=0

The zeros of y^2+2y+5=0 are y=-1+2i,y=-1-2i

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y^{2} + 2 y + 5 = 0

y = - 1 + 2 i, y = - 1 - 2 i

Solution steps

y^{2} + 2 y + 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

Simplify

2^{2} - 4 \cdot 1 \cdot 5 : 4 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = - 1 + 2 i, y = - 1 - 2 i

- 307.6923 x + 2576.92307 = 0

6 x^{2} - 54 x + 84

3 x - 10 < - 25

6 x^{2} + 9 x + 27

- 10 < 2 x + 5 < 7

What are the solutions to the equation y^2+2y+5=0 ?
The solutions to the equation y^2+2y+5=0 are y=-1+2i,y=-1-2i
Find the zeros of y^2+2y+5=0
The zeros of y^2+2y+5=0 are y=-1+2i,y=-1-2i