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

The solutions to the equation 3y^2-y-4=0 are y= 4/3 ,y=-1

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

The zeros of 3y^2-y-4=0 are y= 4/3 ,y=-1

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

y = \frac{4}{3}, y = - 1

Decimal Notation

y = 1.33333 \dots, y = - 1

Solution steps

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 3 (- 4) = 7

y_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 3}, y_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 3}

y = \frac{- ( - 1 ) + 7}{2 \cdot 3} : \frac{4}{3}

y = \frac{- ( - 1 ) - 7}{2 \cdot 3} : - 1

The solutions to the quadratic equation are:

y = \frac{4}{3}, y = - 1

x^{2} + 5 x = 6

\frac{4}{9} \cdot \frac{4}{9}

(x^{2} - 5) = 0

x^{2} + 0 + 0 = 0

2 p^{2} = - 6 p + 1

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