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

The solutions to the equation 8y^2-y+2=0 are y= 1/16+i(3sqrt(7))/(16),y= 1/16-i(3sqrt(7))/(16)

Find the zeros of 8y^2-y+2=0

The zeros of 8y^2-y+2=0 are y= 1/16+i(3sqrt(7))/(16),y= 1/16-i(3sqrt(7))/(16)

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

y = \frac{1}{16} + i \frac{3 7}{16}, y = \frac{1}{16} - i \frac{3 7}{16}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 8 \cdot 2 : 37 i

y_{1, 2} = \frac{- ( - 1 ) \pm 3 7 i}{2 \cdot 8}

Separate the solutions

y_{1} = \frac{- ( - 1 ) + 3 7 i}{2 \cdot 8}, y_{2} = \frac{- ( - 1 ) - 3 7 i}{2 \cdot 8}

y = \frac{- ( - 1 ) + 3 7 i}{2 \cdot 8} : \frac{1}{16} + i \frac{3 7}{16}

y = \frac{- ( - 1 ) - 3 7 i}{2 \cdot 8} : \frac{1}{16} - i \frac{3 7}{16}

The solutions to the quadratic equation are:

y = \frac{1}{16} + i \frac{3 7}{16}, y = \frac{1}{16} - i \frac{3 7}{16}

u^{2} - 4 u = 0

8 x^{2} + 3 = - 4 x

3 n^{2} + n - 310 = 0

x^{2} - 20 x - 12 = - 3

4 x^{2} + 32 x

What are the solutions to the equation 8y^2-y+2=0 ?
The solutions to the equation 8y^2-y+2=0 are y= 1/16+i(3sqrt(7))/(16),y= 1/16-i(3sqrt(7))/(16)
Find the zeros of 8y^2-y+2=0
The zeros of 8y^2-y+2=0 are y= 1/16+i(3sqrt(7))/(16),y= 1/16-i(3sqrt(7))/(16)