What are the solutions to the equation 3y^2+8y-6=0 ?

The solutions to the equation 3y^2+8y-6=0 are y=(-4+sqrt(34))/3 ,y=-(4+sqrt(34))/3

Find the zeros of 3y^2+8y-6=0

The zeros of 3y^2+8y-6=0 are y=(-4+sqrt(34))/3 ,y=-(4+sqrt(34))/3

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3 y^{2} + 8 y - 6 = 0

y = \frac{- 4 + 34}{3}, y = - \frac{4 + 34}{3}

Solution steps

3 y^{2} + 8 y - 6 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 3 ( - 6 )}{2 \cdot 3}

8^{2} - 4 \cdot 3 (- 6) = 234

y_{1, 2} = \frac{- 8 \pm 2 34}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- 8 + 2 34}{2 \cdot 3}, y_{2} = \frac{- 8 - 2 34}{2 \cdot 3}

y = \frac{- 8 + 2 34}{2 \cdot 3} : \frac{- 4 + 34}{3}

y = \frac{- 8 - 2 34}{2 \cdot 3} : - \frac{4 + 34}{3}

The solutions to the quadratic equation are:

y = \frac{- 4 + 34}{3}, y = - \frac{4 + 34}{3}

x^{2} + 3 x - 10

x^{2} - 5 x = - 9

x, x^{2} + y^{2} + 5 z^{2} + 2 x z + 4 yz - 4 = 0

\frac{x ^{2}}{10} - 4 = 12

3 c^{2} - 18 = 0

What are the solutions to the equation 3y^2+8y-6=0 ?
The solutions to the equation 3y^2+8y-6=0 are y=(-4+sqrt(34))/3 ,y=-(4+sqrt(34))/3
Find the zeros of 3y^2+8y-6=0
The zeros of 3y^2+8y-6=0 are y=(-4+sqrt(34))/3 ,y=-(4+sqrt(34))/3