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

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

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

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

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m^{2} - m + 1 = 0

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

Solution steps

m^{2} - m + 1 = 0

Solve with the quadratic formula

m_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

4 - 7 x + 10 = 15 - 6 x - 8

2 ∣ t - 3 ∣ - 3 = 7

(t - 7)^{2} = (t + 7)^{2} - 112

3 (x - 1) - 2 (x + 1) + 4 (x - 2) = 0

4 - 3 x = 2 + 5 x

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