What are the solutions to the equation 4v^2=2-2v ?

The solutions to the equation 4v^2=2-2v are v= 1/2 ,v=-1

Find the zeros of 4v^2=2-2v

The zeros of 4v^2=2-2v are v= 1/2 ,v=-1

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4 v^{2} = 2 - 2 v

v = \frac{1}{2}, v = - 1

Decimal Notation

v = 0.5, v = - 1

Solution steps

4 v^{2} = 2 - 2 v

Move

2 v

to the left side

4 v^{2} + 2 v = 2

Move

2

to the left side

4 v^{2} + 2 v - 2 = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 4 (- 2) = 6

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

Separate the solutions

v_{1} = \frac{- 2 + 6}{2 \cdot 4}, v_{2} = \frac{- 2 - 6}{2 \cdot 4}

v = \frac{- 2 + 6}{2 \cdot 4} : \frac{1}{2}

v = \frac{- 2 - 6}{2 \cdot 4} : - 1

The solutions to the quadratic equation are:

v = \frac{1}{2}, v = - 1

4.9 x^{2} - 39.2 x + 58.8 = 0

5 x^{2} + 12 x = 0

2 x^{2} + x^{2} = 100

x^{2} + 14 x + 43 = 0

x (x - 9) = 5

What are the solutions to the equation 4v^2=2-2v ?
The solutions to the equation 4v^2=2-2v are v= 1/2 ,v=-1
Find the zeros of 4v^2=2-2v
The zeros of 4v^2=2-2v are v= 1/2 ,v=-1