What are the solutions to the equation 6x^2-x+8=0 ?

The solutions to the equation 6x^2-x+8=0 are x= 1/12+i(sqrt(191))/(12),x= 1/12-i(sqrt(191))/(12)

Find the zeros of 6x^2-x+8=0

The zeros of 6x^2-x+8=0 are x= 1/12+i(sqrt(191))/(12),x= 1/12-i(sqrt(191))/(12)

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6 x^{2} - x + 8 = 0

x = \frac{1}{12} + i \frac{191}{12}, x = \frac{1}{12} - i \frac{191}{12}

Solution steps

6 x^{2} - x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 6 \cdot 8}{2 \cdot 6}

Simplify

(- 1)^{2} - 4 \cdot 6 \cdot 8 : 191 i

x_{1, 2} = \frac{- ( - 1 ) \pm 191 i}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 191 i}{2 \cdot 6}, x_{2} = \frac{- ( - 1 ) - 191 i}{2 \cdot 6}

x = \frac{- ( - 1 ) + 191 i}{2 \cdot 6} : \frac{1}{12} + i \frac{191}{12}

x = \frac{- ( - 1 ) - 191 i}{2 \cdot 6} : \frac{1}{12} - i \frac{191}{12}

The solutions to the quadratic equation are:

x = \frac{1}{12} + i \frac{191}{12}, x = \frac{1}{12} - i \frac{191}{12}

4 x^{3} - 2 x^{2} - 12 x

\frac{1}{\frac{3}{3}} = \frac{x}{8}

5^{\frac{1}{4}} \cdot 5^{\frac{7}{4}}

10 z^{2} - 9 z + 2 = 0

What are the solutions to the equation 6x^2-x+8=0 ?
The solutions to the equation 6x^2-x+8=0 are x= 1/12+i(sqrt(191))/(12),x= 1/12-i(sqrt(191))/(12)
Find the zeros of 6x^2-x+8=0
The zeros of 6x^2-x+8=0 are x= 1/12+i(sqrt(191))/(12),x= 1/12-i(sqrt(191))/(12)