What are the solutions to the equation 3y^2+y-1=0 ?

The solutions to the equation 3y^2+y-1=0 are y=(-1+sqrt(13))/6 ,y=(-1-sqrt(13))/6

Find the zeros of 3y^2+y-1=0

The zeros of 3y^2+y-1=0 are y=(-1+sqrt(13))/6 ,y=(-1-sqrt(13))/6

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3 y^{2} + y - 1 = 0

y = \frac{- 1 + 13}{6}, y = \frac{- 1 - 13}{6}

Decimal Notation

y = 0.43425 \dots, y = - 0.76759 \dots

Solution steps

3 y^{2} + y - 1 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 3 (- 1) = 13

y_{1, 2} = \frac{- 1 \pm 13}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- 1 + 13}{2 \cdot 3}, y_{2} = \frac{- 1 - 13}{2 \cdot 3}

y = \frac{- 1 + 13}{2 \cdot 3} : \frac{- 1 + 13}{6}

y = \frac{- 1 - 13}{2 \cdot 3} : \frac{- 1 - 13}{6}

The solutions to the quadratic equation are:

y = \frac{- 1 + 13}{6}, y = \frac{- 1 - 13}{6}

x, x^{2} - 5 = 0

5 u^{2} + 7 u = - 2

(x - 5)^{2} - 8 = - 26

p^{2} + 2 p - 1 = 0

5 w^{2} - 9 w = 2

What are the solutions to the equation 3y^2+y-1=0 ?
The solutions to the equation 3y^2+y-1=0 are y=(-1+sqrt(13))/6 ,y=(-1-sqrt(13))/6
Find the zeros of 3y^2+y-1=0
The zeros of 3y^2+y-1=0 are y=(-1+sqrt(13))/6 ,y=(-1-sqrt(13))/6