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

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

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

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

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

v = \frac{3 + 2 3}{3}, v = \frac{3 - 2 3}{3}

Solution steps

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

Move

3

to the left side

3 v^{2} - 6 v - 1 = 0

Solve with the quadratic formula

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

(- 6)^{2} - 4 \cdot 3 (- 1) = 43

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

Separate the solutions

v_{1} = \frac{- ( - 6 ) + 4 3}{2 \cdot 3}, v_{2} = \frac{- ( - 6 ) - 4 3}{2 \cdot 3}

v = \frac{- ( - 6 ) + 4 3}{2 \cdot 3} : \frac{3 + 2 3}{3}

v = \frac{- ( - 6 ) - 4 3}{2 \cdot 3} : \frac{3 - 2 3}{3}

The solutions to the quadratic equation are:

v = \frac{3 + 2 3}{3}, v = \frac{3 - 2 3}{3}

x^{2} - 4 = - 4 x

2 x - 3 (5 x - 1) < 3 - (13 x + 8)

x^{2} - 64 x = 0

2 x + 16 < x + 25

x^{3} - 4 x^{2} - 16 x + 64 < 0

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