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

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

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

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

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1 x^{2} + 1 x + 1 = 0

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

Solution steps

1 \cdot x^{2} + 1 \cdot x + 1 = 0

Refine

x^{2} + x + 1 = 0

Solve with the quadratic formula

x_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

26981.75 \cdot (1 + 0.16)^{15}

x^{4} - 13 x^{2} + 36

8 \cdot 3^{3}

(p^{- \frac{2}{3}} q^{\frac{1}{2}})^{3}

x^{2} - 6 x + 24 = 0

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