What are the solutions to the equation 3y^2-y+1=0 ?

The solutions to the equation 3y^2-y+1=0 are y= 1/6+i(sqrt(11))/6 ,y= 1/6-i(sqrt(11))/6

Find the zeros of 3y^2-y+1=0

The zeros of 3y^2-y+1=0 are y= 1/6+i(sqrt(11))/6 ,y= 1/6-i(sqrt(11))/6

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3 y^{2} - y + 1 = 0

y = \frac{1}{6} + i \frac{11}{6}, y = \frac{1}{6} - i \frac{11}{6}

Solution steps

3 y^{2} - y + 1 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

Simplify

(- 1)^{2} - 4 \cdot 3 \cdot 1 : 11 i

y_{1, 2} = \frac{- ( - 1 ) \pm 11 i}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 1 ) + 11 i}{2 \cdot 3}, y_{2} = \frac{- ( - 1 ) - 11 i}{2 \cdot 3}

y = \frac{- ( - 1 ) + 11 i}{2 \cdot 3} : \frac{1}{6} + i \frac{11}{6}

y = \frac{- ( - 1 ) - 11 i}{2 \cdot 3} : \frac{1}{6} - i \frac{11}{6}

The solutions to the quadratic equation are:

y = \frac{1}{6} + i \frac{11}{6}, y = \frac{1}{6} - i \frac{11}{6}

(a x^{2} - 5 x + 1) + (b x - 16 x^{2} - 9) = (- 14 x^{2} - 10 x + c)

5 x^{2} - 10 x + 9 = 3

- 2 x^{2} + 11 x + 7 = 10 + 4 x

x^{2} + 10 x - 11 = 0

x^{2} - (3 \cdot (- 4)) = 100 + 3.5 - (\frac{- 24}{2})

What are the solutions to the equation 3y^2-y+1=0 ?
The solutions to the equation 3y^2-y+1=0 are y= 1/6+i(sqrt(11))/6 ,y= 1/6-i(sqrt(11))/6
Find the zeros of 3y^2-y+1=0
The zeros of 3y^2-y+1=0 are y= 1/6+i(sqrt(11))/6 ,y= 1/6-i(sqrt(11))/6