What are the solutions to the equation 6v^2+3=-2v ?

The solutions to the equation 6v^2+3=-2v are v=-1/6+i(sqrt(17))/6 ,v=-1/6-i(sqrt(17))/6

Find the zeros of 6v^2+3=-2v

The zeros of 6v^2+3=-2v are v=-1/6+i(sqrt(17))/6 ,v=-1/6-i(sqrt(17))/6

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6 v^{2} + 3 = - 2 v

v = - \frac{1}{6} + i \frac{17}{6}, v = - \frac{1}{6} - i \frac{17}{6}

Solution steps

6 v^{2} + 3 = - 2 v

Move

2 v

to the left side

6 v^{2} + 3 + 2 v = 0

Write in the standard form

a x^{2} + b x + c = 0

6 v^{2} + 2 v + 3 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 6 \cdot 3}{2 \cdot 6}

Simplify

2^{2} - 4 \cdot 6 \cdot 3 : 217 i

v_{1, 2} = \frac{- 2 \pm 2 17 i}{2 \cdot 6}

Separate the solutions

v_{1} = \frac{- 2 + 2 17 i}{2 \cdot 6}, v_{2} = \frac{- 2 - 2 17 i}{2 \cdot 6}

v = \frac{- 2 + 2 17 i}{2 \cdot 6} : - \frac{1}{6} + i \frac{17}{6}

v = \frac{- 2 - 2 17 i}{2 \cdot 6} : - \frac{1}{6} - i \frac{17}{6}

The solutions to the quadratic equation are:

v = - \frac{1}{6} + i \frac{17}{6}, v = - \frac{1}{6} - i \frac{17}{6}

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

2 a^{2} - 288 = 0

x (4 x - 19) = 30

x^{2} - 2 a x - b = 0

45 y + 7 y^{2} = 28

What are the solutions to the equation 6v^2+3=-2v ?
The solutions to the equation 6v^2+3=-2v are v=-1/6+i(sqrt(17))/6 ,v=-1/6-i(sqrt(17))/6
Find the zeros of 6v^2+3=-2v
The zeros of 6v^2+3=-2v are v=-1/6+i(sqrt(17))/6 ,v=-1/6-i(sqrt(17))/6