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

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

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

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

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t^{2} + t + 1 = 0

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

Solution steps

t^{2} + t + 1 = 0

Solve with the quadratic formula

t_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

(\frac{L}{2}) + (\frac{L}{2})

8 x^{3} + y^{3}

\frac{10 \cdot 7}{17.5}

4 (3 x + 4) - 48 = 6 x - 2

3 x + 8 = 11

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