What are the solutions to the equation 3y^2+2y-2=0 ?

The solutions to the equation 3y^2+2y-2=0 are y=(-1+sqrt(7))/3 ,y=-(1+sqrt(7))/3

Find the zeros of 3y^2+2y-2=0

The zeros of 3y^2+2y-2=0 are y=(-1+sqrt(7))/3 ,y=-(1+sqrt(7))/3

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3 y^{2} + 2 y - 2 = 0

y = \frac{- 1 + 7}{3}, y = - \frac{1 + 7}{3}

Decimal Notation

y = 0.54858 \dots, y = - 1.21525 \dots

Solution steps

3 y^{2} + 2 y - 2 = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 3 (- 2) = 27

y_{1, 2} = \frac{- 2 \pm 2 7}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- 2 + 2 7}{2 \cdot 3}, y_{2} = \frac{- 2 - 2 7}{2 \cdot 3}

y = \frac{- 2 + 2 7}{2 \cdot 3} : \frac{- 1 + 7}{3}

y = \frac{- 2 - 2 7}{2 \cdot 3} : - \frac{1 + 7}{3}

The solutions to the quadratic equation are:

y = \frac{- 1 + 7}{3}, y = - \frac{1 + 7}{3}

x, z = y^{2} - x^{2} + 2

x (x + 1 0^{- 8}) = 1 0^{- 14}

- 27 = 4 x^{2} - 24 x

x, 21.8 x^{2} - 202 x - 60 = 0

x, 4 x^{2} + 8 y^{2} - 4 z^{2} = 0

What are the solutions to the equation 3y^2+2y-2=0 ?
The solutions to the equation 3y^2+2y-2=0 are y=(-1+sqrt(7))/3 ,y=-(1+sqrt(7))/3
Find the zeros of 3y^2+2y-2=0
The zeros of 3y^2+2y-2=0 are y=(-1+sqrt(7))/3 ,y=-(1+sqrt(7))/3