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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

(x^{2} - 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}

x, 4 x^{2} - 24 x y + 11 y^{2} + 56 x - 58 y + 95 = 0

t^{2} - 2 t + 1 = 0

5 x^{2} - 100 x + 255 = 0

(x + 4)^{2} - 16 = 100

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

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