What are the solutions to the equation 6x^2-x+1=0 ?

The solutions to the equation 6x^2-x+1=0 are x= 1/12+i(sqrt(23))/(12),x= 1/12-i(sqrt(23))/(12)

Find the zeros of 6x^2-x+1=0

The zeros of 6x^2-x+1=0 are x= 1/12+i(sqrt(23))/(12),x= 1/12-i(sqrt(23))/(12)

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

x = \frac{1}{12} + i \frac{23}{12}, x = \frac{1}{12} - i \frac{23}{12}

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

Simplify

(- 1)^{2} - 4 \cdot 6 \cdot 1 : 23 i

x_{1, 2} = \frac{- ( - 1 ) \pm 23 i}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 23 i}{2 \cdot 6}, x_{2} = \frac{- ( - 1 ) - 23 i}{2 \cdot 6}

x = \frac{- ( - 1 ) + 23 i}{2 \cdot 6} : \frac{1}{12} + i \frac{23}{12}

x = \frac{- ( - 1 ) - 23 i}{2 \cdot 6} : \frac{1}{12} - i \frac{23}{12}

The solutions to the quadratic equation are:

x = \frac{1}{12} + i \frac{23}{12}, x = \frac{1}{12} - i \frac{23}{12}

r^{2} - 2 r + 4 = 0

s^{2} - 8 s + 20

4 x^{2} + 12 x + 9 = 49

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

x, x^{2} + y^{2} - 4 x - 2 y + 36 = 0

What are the solutions to the equation 6x^2-x+1=0 ?
The solutions to the equation 6x^2-x+1=0 are x= 1/12+i(sqrt(23))/(12),x= 1/12-i(sqrt(23))/(12)
Find the zeros of 6x^2-x+1=0
The zeros of 6x^2-x+1=0 are x= 1/12+i(sqrt(23))/(12),x= 1/12-i(sqrt(23))/(12)