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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 y^{2} = - y + 2

y = \frac{2}{3}, y = - 1

Solution steps

3 y^{2} = - y + 2

Move

2

to the left side

3 y^{2} - 2 = - y

Move

y

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 3 (- 2) = 5

y_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- 1 + 5}{2 \cdot 3}, y_{2} = \frac{- 1 - 5}{2 \cdot 3}

y = \frac{- 1 + 5}{2 \cdot 3} : \frac{2}{3}

y = \frac{- 1 - 5}{2 \cdot 3} : - 1

The solutions to the quadratic equation are:

y = \frac{2}{3}, y = - 1

x^{3} - 4 x^{2} + 25 x - 100

\frac{2}{5} + 1

6 x^{2} - 25 x - 25 = 0

\frac{1}{2} - \frac{1}{1}

x^{3} - 6 x^{2} + 32

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