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

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

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

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

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3 y^{2} - y = 2

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

Solution steps

3 y^{2} - y = 2

Move

2

to the left side

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} : 1

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

The solutions to the quadratic equation are:

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

2^{22}

3 x = - 18

∣ x ∣ = 17

\frac{5 b - 10}{b ^{2} + 5 b - 14}

(w + 1) (w - 1) = w (w - 4)

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