What are the solutions to the equation 4v^2-4=-3v+3v^2 ?

The solutions to the equation 4v^2-4=-3v+3v^2 are v=1,v=-4

Find the zeros of 4v^2-4=-3v+3v^2

The zeros of 4v^2-4=-3v+3v^2 are v=1,v=-4

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4 v^{2} - 4 = - 3 v + 3 v^{2}

v = 1, v = - 4

Solution steps

4 v^{2} - 4 = - 3 v + 3 v^{2}

Move

3 v^{2}

to the left side

v^{2} - 4 = - 3 v

Move

3 v

to the left side

v^{2} - 4 + 3 v = 0

Write in the standard form

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

v^{2} + 3 v - 4 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 4) = 5

v_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- 3 + 5}{2 \cdot 1}, v_{2} = \frac{- 3 - 5}{2 \cdot 1}

v = \frac{- 3 + 5}{2 \cdot 1} : 1

v = \frac{- 3 - 5}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

v = 1, v = - 4

8 a^{2} + 28 a = - 20

4 z^{2} = 144

x^{2} - 3 x + \frac{37}{16} = 0

8 x^{2} - 19 = 15

3 x^{2} - 2.2 + 5 x = 0

What are the solutions to the equation 4v^2-4=-3v+3v^2 ?
The solutions to the equation 4v^2-4=-3v+3v^2 are v=1,v=-4
Find the zeros of 4v^2-4=-3v+3v^2
The zeros of 4v^2-4=-3v+3v^2 are v=1,v=-4