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

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

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

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

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k^{2} + k + 1 = 0

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

Solution steps

k^{2} + k + 1 = 0

Solve with the quadratic formula

k_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

(- 3) (2 x - 2) < (- 3) (2)

\frac{5}{12 x ^{4} y} - \frac{1}{5 x ^{2} y ^{3}}

5 ∣ x + 10 ∣ + 4 < 14

2 - 5 z \geq - 13

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

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