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

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

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

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

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w^{2} - w + 1 = 0

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

Solution steps

w^{2} - w + 1 = 0

Solve with the quadratic formula

w_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

n^{2} = - 7 n - 10

(x + 4) (x + 7) = 0

x^{2} + 4 x - 35 = 0

(5 x - 8) (x + 2) = (x - 1) (7 x - 4)

x^{2} - 12 + 35 = 0

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