What are the solutions to the equation 3y^2+7y-7=0 ?

The solutions to the equation 3y^2+7y-7=0 are y=(-7+sqrt(133))/6 ,y=(-7-sqrt(133))/6

Find the zeros of 3y^2+7y-7=0

The zeros of 3y^2+7y-7=0 are y=(-7+sqrt(133))/6 ,y=(-7-sqrt(133))/6

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3 y^{2} + 7 y - 7 = 0

y = \frac{- 7 + 133}{6}, y = \frac{- 7 - 133}{6}

Decimal Notation

y = 0.75542 \dots, y = - 3.08876 \dots

Solution steps

3 y^{2} + 7 y - 7 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 3 ( - 7 )}{2 \cdot 3}

7^{2} - 4 \cdot 3 (- 7) = 133

y_{1, 2} = \frac{- 7 \pm 133}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- 7 + 133}{2 \cdot 3}, y_{2} = \frac{- 7 - 133}{2 \cdot 3}

y = \frac{- 7 + 133}{2 \cdot 3} : \frac{- 7 + 133}{6}

y = \frac{- 7 - 133}{2 \cdot 3} : \frac{- 7 - 133}{6}

The solutions to the quadratic equation are:

y = \frac{- 7 + 133}{6}, y = \frac{- 7 - 133}{6}

2 x^{2} + 10 x - 15 = 0

- 2 x^{2} + 12 x - 13 = 0

2 x = - 2 x^{2} + 8

17 w^{2} + 4 w - 7 = 2 w^{2} + 6 w + 1

λ^{2} - 9 λ + 18 = 0

What are the solutions to the equation 3y^2+7y-7=0 ?
The solutions to the equation 3y^2+7y-7=0 are y=(-7+sqrt(133))/6 ,y=(-7-sqrt(133))/6
Find the zeros of 3y^2+7y-7=0
The zeros of 3y^2+7y-7=0 are y=(-7+sqrt(133))/6 ,y=(-7-sqrt(133))/6