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

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

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

The zeros of 3y^2+2y-1=0 are y= 1/3 ,y=-1

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

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

Decimal Notation

y = 0.33333 \dots, y = - 1

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

y_{1} = \frac{- 2 + 4}{2 \cdot 3}, y_{2} = \frac{- 2 - 4}{2 \cdot 3}

y = \frac{- 2 + 4}{2 \cdot 3} : \frac{1}{3}

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

The solutions to the quadratic equation are:

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

3969

2 n^{2} + 17 n - 30

144 - 121 x^{2}

\frac{17}{1 - i}

(m^{\frac{5}{4}} n^{- \frac{4}{5}})^{\frac{7}{3}}

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