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

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

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

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

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

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

Solution steps

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}

x, 2 x \cdot (6 x - 2 y^{2}) + 3 y^{2} (x - 4) = 8

x^{2} + (m - 1) x + 2 m - 6 = 0

x, x^{2} + y^{3} + 2 x - 8 y - 113 = 130

\frac{( x ^{2} )}{9} - \frac{x}{2} = \frac{1}{3} + \frac{x}{9}

x, x^{2} + y^{2} = b^{2}

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