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

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

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

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

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

y = \frac{1}{3}, y = 0

Decimal Notation

y = 0.33333 \dots, y = 0

Solution steps

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

Refine

3 y^{2} - y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

(- 1)^{2} - 4 \cdot 3 \cdot 0 = 1

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

Separate the solutions

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

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

y = \frac{- ( - 1 ) - 1}{2 \cdot 3} : 0

The solutions to the quadratic equation are:

y = \frac{1}{3}, y = 0

\frac{3}{4} + \frac{1}{5} x - \frac{1}{3} = \frac{4}{5} x

9 x - 5 = 15 x - 23

0.6 (10 n + 25) = 10 + 5 n

24

\frac{1}{12} (4) (2.5)^{3}

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