What are the solutions to the equation y^2+2y+3=0 ?

The solutions to the equation y^2+2y+3=0 are y=-1+sqrt(2)i,y=-1-sqrt(2)i

Find the zeros of y^2+2y+3=0

The zeros of y^2+2y+3=0 are y=-1+sqrt(2)i,y=-1-sqrt(2)i

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y^{2} + 2 y + 3 = 0

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

Solution steps

y^{2} + 2 y + 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

Simplify

2^{2} - 4 \cdot 1 \cdot 3 : 22 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

2 x^{2} - 3 x - 5 = 0

(3 x + 2) (x - 6) = 0

7 - 9 x + 4 x^{2} = 0

s^{2} + 40 s + 1600 = 0

8 - y^{2} = 4 y + 12

What are the solutions to the equation y^2+2y+3=0 ?
The solutions to the equation y^2+2y+3=0 are y=-1+sqrt(2)i,y=-1-sqrt(2)i
Find the zeros of y^2+2y+3=0
The zeros of y^2+2y+3=0 are y=-1+sqrt(2)i,y=-1-sqrt(2)i