What are the solutions to the equation 8v^2+v=-9 ?

The solutions to the equation 8v^2+v=-9 are v=-1/16+i(sqrt(287))/(16),v=-1/16-i(sqrt(287))/(16)

Find the zeros of 8v^2+v=-9

The zeros of 8v^2+v=-9 are v=-1/16+i(sqrt(287))/(16),v=-1/16-i(sqrt(287))/(16)

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8 v^{2} + v = - 9

v = - \frac{1}{16} + i \frac{287}{16}, v = - \frac{1}{16} - i \frac{287}{16}

Solution steps

8 v^{2} + v = - 9

Move

9

to the left side

8 v^{2} + v + 9 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 8 \cdot 9}{2 \cdot 8}

Simplify

1^{2} - 4 \cdot 8 \cdot 9 : 287 i

v_{1, 2} = \frac{- 1 \pm 287 i}{2 \cdot 8}

Separate the solutions

v_{1} = \frac{- 1 + 287 i}{2 \cdot 8}, v_{2} = \frac{- 1 - 287 i}{2 \cdot 8}

v = \frac{- 1 + 287 i}{2 \cdot 8} : - \frac{1}{16} + i \frac{287}{16}

v = \frac{- 1 - 287 i}{2 \cdot 8} : - \frac{1}{16} - i \frac{287}{16}

The solutions to the quadratic equation are:

v = - \frac{1}{16} + i \frac{287}{16}, v = - \frac{1}{16} - i \frac{287}{16}

8 + ∣ 4 x + 1 ∣ = 8

2 y + 42 + 2 y = 6

7 x - 12 = 23

5 - 7 (a + 2) = 7 + a

a^{6} + b^{6}

What are the solutions to the equation 8v^2+v=-9 ?
The solutions to the equation 8v^2+v=-9 are v=-1/16+i(sqrt(287))/(16),v=-1/16-i(sqrt(287))/(16)
Find the zeros of 8v^2+v=-9
The zeros of 8v^2+v=-9 are v=-1/16+i(sqrt(287))/(16),v=-1/16-i(sqrt(287))/(16)