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

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

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

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

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

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

Decimal Notation

y = - 1, y = 0.66666 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

0 = 16 - 0.01 s^{2}

x (3 x - 14) = 5

15 = \frac{1}{3} q^{2}

z^{2} - 1.8994 z + 0.9607 = 0

x^{2} + a x + 4 = 0

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