What are the solutions to the equation (x^2-x+1)/4 =0 ?

The solutions to the equation (x^2-x+1)/4 =0 are x= 1/2+i(sqrt(3))/2 ,x= 1/2-i(sqrt(3))/2

Find the zeros of (x^2-x+1)/4 =0

The zeros of (x^2-x+1)/4 =0 are x= 1/2+i(sqrt(3))/2 ,x= 1/2-i(sqrt(3))/2

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\frac{x ^{2} - x + 1}{4} = 0

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

Solution steps

\frac{x ^{2} - x + 1}{4} = 0

Multiply both sides by

4

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} + 23 x + 76 = 0

8 x^{2} + 3 = 0

9 x^{2} + 9 x = 7

196 q - q^{2} = 0

3 x^{2} = 14 x + 5

What are the solutions to the equation (x^2-x+1)/4 =0 ?
The solutions to the equation (x^2-x+1)/4 =0 are x= 1/2+i(sqrt(3))/2 ,x= 1/2-i(sqrt(3))/2
Find the zeros of (x^2-x+1)/4 =0
The zeros of (x^2-x+1)/4 =0 are x= 1/2+i(sqrt(3))/2 ,x= 1/2-i(sqrt(3))/2