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

The solutions to the equation 3y^2-2y-4=0 are y=(1+sqrt(13))/3 ,y=(1-sqrt(13))/3

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

The zeros of 3y^2-2y-4=0 are y=(1+sqrt(13))/3 ,y=(1-sqrt(13))/3

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

y = \frac{1 + 13}{3}, y = \frac{1 - 13}{3}

Decimal Notation

y = 1.53518 \dots, y = - 0.86851 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 3 (- 4) = 213

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = \frac{1 + 13}{3}, y = \frac{1 - 13}{3}

v^{2} - 2 v - 35 = 0

- 108 = - 16 t^{2} - 22 t

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

w^{2} = 5 (w - 4)

6 x^{2} - 7 x + 6 = 0

What are the solutions to the equation 3y^2-2y-4=0 ?
The solutions to the equation 3y^2-2y-4=0 are y=(1+sqrt(13))/3 ,y=(1-sqrt(13))/3
Find the zeros of 3y^2-2y-4=0
The zeros of 3y^2-2y-4=0 are y=(1+sqrt(13))/3 ,y=(1-sqrt(13))/3