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

The solutions to the equation 2y^2+2y+3=0 are y=-1/2+i(sqrt(5))/2 ,y=-1/2-i(sqrt(5))/2

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

The zeros of 2y^2+2y+3=0 are y=-1/2+i(sqrt(5))/2 ,y=-1/2-i(sqrt(5))/2

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

y = - \frac{1}{2} + i \frac{5}{2}, y = - \frac{1}{2} - i \frac{5}{2}

Solution steps

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

Solve with the quadratic formula

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

Simplify

2^{2} - 4 \cdot 2 \cdot 3 : 25 i

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

Separate the solutions

y_{1} = \frac{- 2 + 2 5 i}{2 \cdot 2}, y_{2} = \frac{- 2 - 2 5 i}{2 \cdot 2}

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

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

The solutions to the quadratic equation are:

y = - \frac{1}{2} + i \frac{5}{2}, y = - \frac{1}{2} - i \frac{5}{2}

0.534 = 4.9 t^{2} + 1.08 t

x^{2} + 3 x - 6 = 0

4 = \frac{x}{- 4} + 8

12 (x - 1) = 8 (x + 1)

(- 1)^{- \frac{1}{2}}

What are the solutions to the equation 2y^2+2y+3=0 ?
The solutions to the equation 2y^2+2y+3=0 are y=-1/2+i(sqrt(5))/2 ,y=-1/2-i(sqrt(5))/2
Find the zeros of 2y^2+2y+3=0
The zeros of 2y^2+2y+3=0 are y=-1/2+i(sqrt(5))/2 ,y=-1/2-i(sqrt(5))/2