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

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

Find the zeros of 6w^2+w+1=0

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

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6 w^{2} + w + 1 = 0

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

Solution steps

6 w^{2} + w + 1 = 0

Solve with the quadratic formula

w_{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

w_{1, 2} = \frac{- 1 \pm 23 i}{2 \cdot 6}

Separate the solutions

w_{1} = \frac{- 1 + 23 i}{2 \cdot 6}, w_{2} = \frac{- 1 - 23 i}{2 \cdot 6}

w = \frac{- 1 + 23 i}{2 \cdot 6} : - \frac{1}{12} + i \frac{23}{12}

w = \frac{- 1 - 23 i}{2 \cdot 6} : - \frac{1}{12} - i \frac{23}{12}

The solutions to the quadratic equation are:

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

(\frac{15}{2})^{2}

(1 - 2)^{2}

18 - 3 x - 4 = 3 x - 4

(1 - x)^{3}

\frac{1}{3} (2 x + 3) = - 9

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