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

The solutions to the equation 6y^2-y+6=0 are y= 1/12+i(sqrt(143))/(12),y= 1/12-i(sqrt(143))/(12)

Find the zeros of 6y^2-y+6=0

The zeros of 6y^2-y+6=0 are y= 1/12+i(sqrt(143))/(12),y= 1/12-i(sqrt(143))/(12)

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6 y^{2} - y + 6 = 0

y = \frac{1}{12} + i \frac{143}{12}, y = \frac{1}{12} - i \frac{143}{12}

Solution steps

6 y^{2} - y + 6 = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 6 \cdot 6 : 143 i

y_{1, 2} = \frac{- ( - 1 ) \pm 143 i}{2 \cdot 6}

Separate the solutions

y_{1} = \frac{- ( - 1 ) + 143 i}{2 \cdot 6}, y_{2} = \frac{- ( - 1 ) - 143 i}{2 \cdot 6}

y = \frac{- ( - 1 ) + 143 i}{2 \cdot 6} : \frac{1}{12} + i \frac{143}{12}

y = \frac{- ( - 1 ) - 143 i}{2 \cdot 6} : \frac{1}{12} - i \frac{143}{12}

The solutions to the quadratic equation are:

y = \frac{1}{12} + i \frac{143}{12}, y = \frac{1}{12} - i \frac{143}{12}

25 y^{2} + 40 y = - 16

x^{2} - 6 x + 12 = 41 + 12

\frac{2}{5} x^{2} - x = \frac{1}{5}

r, A = π r^{2} h

x, 3 x^{2} + 5 x - 1 = 0

What are the solutions to the equation 6y^2-y+6=0 ?
The solutions to the equation 6y^2-y+6=0 are y= 1/12+i(sqrt(143))/(12),y= 1/12-i(sqrt(143))/(12)
Find the zeros of 6y^2-y+6=0
The zeros of 6y^2-y+6=0 are y= 1/12+i(sqrt(143))/(12),y= 1/12-i(sqrt(143))/(12)