What are the solutions to the equation y^2+y+1=0 ?

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

Find the zeros of y^2+y+1=0

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

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y^{2} + y + 1 = 0

y = - \frac{1}{2} + i \frac{3}{2}, y = - \frac{1}{2} - i \frac{3}{2}

Solution steps

y^{2} + y + 1 = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

y_{1} = \frac{- 1 + 3 i}{2 \cdot 1}, y_{2} = \frac{- 1 - 3 i}{2 \cdot 1}

y = \frac{- 1 + 3 i}{2 \cdot 1} : - \frac{1}{2} + i \frac{3}{2}

y = \frac{- 1 - 3 i}{2 \cdot 1} : - \frac{1}{2} - i \frac{3}{2}

The solutions to the quadratic equation are:

y = - \frac{1}{2} + i \frac{3}{2}, y = - \frac{1}{2} - i \frac{3}{2}

2 x^{2} - 4 x + 8 = 0

x^{2} + \frac{1}{3} x + 5

- 2 x^{2} + 6 = 0

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

p^{2} - 16 p + 13 = - 3 p^{2}

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