What are the solutions to the equation 3y^2+6y-4=0 ?

The solutions to the equation 3y^2+6y-4=0 are y=(-3+sqrt(21))/3 ,y=-(3+sqrt(21))/3

Find the zeros of 3y^2+6y-4=0

The zeros of 3y^2+6y-4=0 are y=(-3+sqrt(21))/3 ,y=-(3+sqrt(21))/3

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3 y^{2} + 6 y - 4 = 0

y = \frac{- 3 + 21}{3}, y = - \frac{3 + 21}{3}

Solution steps

3 y^{2} + 6 y - 4 = 0

Solve with the quadratic formula

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

6^{2} - 4 \cdot 3 (- 4) = 221

y_{1, 2} = \frac{- 6 \pm 2 21}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- 6 + 2 21}{2 \cdot 3}, y_{2} = \frac{- 6 - 2 21}{2 \cdot 3}

y = \frac{- 6 + 2 21}{2 \cdot 3} : \frac{- 3 + 21}{3}

y = \frac{- 6 - 2 21}{2 \cdot 3} : - \frac{3 + 21}{3}

The solutions to the quadratic equation are:

y = \frac{- 3 + 21}{3}, y = - \frac{3 + 21}{3}

a^{2} - 4 ab - 45 b^{2}

12 \cdot \frac{1}{4}

x y^{2} - 36 x - 5 y^{2} + 180

8 x < 16

6 x^{2} = 30

What are the solutions to the equation 3y^2+6y-4=0 ?
The solutions to the equation 3y^2+6y-4=0 are y=(-3+sqrt(21))/3 ,y=-(3+sqrt(21))/3
Find the zeros of 3y^2+6y-4=0
The zeros of 3y^2+6y-4=0 are y=(-3+sqrt(21))/3 ,y=-(3+sqrt(21))/3