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

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

Find the zeros of s^2-s+1=0

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

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s^{2} - s + 1 = 0

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

Solution steps

s^{2} - s + 1 = 0

Solve with the quadratic formula

s_{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

s_{1, 2} = \frac{- ( - 1 ) \pm 3 i}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- ( - 1 ) + 3 i}{2 \cdot 1}, s_{2} = \frac{- ( - 1 ) - 3 i}{2 \cdot 1}

s = \frac{- ( - 1 ) + 3 i}{2 \cdot 1} : \frac{1}{2} + i \frac{3}{2}

s = \frac{- ( - 1 ) - 3 i}{2 \cdot 1} : \frac{1}{2} - i \frac{3}{2}

The solutions to the quadratic equation are:

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

x^{2} + 4^{2} = 6^{2}

32 x^{2} - 6 x + 9 = 0

x^{2} - 3 x - 1

z^{2} - 12 z + 35 = 0

2 x (2 x + 1) = 5

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