What are the solutions to the equation -3y^2=6y+2 ?

The solutions to the equation -3y^2=6y+2 are y=-(3+sqrt(3))/3 ,y=-(3-sqrt(3))/3

Find the zeros of -3y^2=6y+2

The zeros of -3y^2=6y+2 are y=-(3+sqrt(3))/3 ,y=-(3-sqrt(3))/3

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- 3 y^{2} = 6 y + 2

y = - \frac{3 + 3}{3}, y = - \frac{3 - 3}{3}

Decimal Notation

y = - 1.57735 \dots, y = - 0.42264 \dots

Solution steps

- 3 y^{2} = 6 y + 2

Move

2

to the left side

- 3 y^{2} - 2 = 6 y

Move

6 y

to the left side

- 3 y^{2} - 2 - 6 y = 0

Write in the standard form

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

- 3 y^{2} - 6 y - 2 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 3 ) ( - 2 )}{2 ( - 3 )}

(- 6)^{2} - 4 (- 3) (- 2) = 23

y_{1, 2} = \frac{- ( - 6 ) \pm 2 3}{2 ( - 3 )}

Separate the solutions

y_{1} = \frac{- ( - 6 ) + 2 3}{2 ( - 3 )}, y_{2} = \frac{- ( - 6 ) - 2 3}{2 ( - 3 )}

y = \frac{- ( - 6 ) + 2 3}{2 ( - 3 )} : - \frac{3 + 3}{3}

y = \frac{- ( - 6 ) - 2 3}{2 ( - 3 )} : - \frac{3 - 3}{3}

The solutions to the quadratic equation are:

y = - \frac{3 + 3}{3}, y = - \frac{3 - 3}{3}

\frac{3}{2} \cdot 4

- 3 z^{2} - 2 z + 6 = 0

\frac{x ^{- 1} + y ^{- 1}}{( x + y ) ^{- 1}}

- 6 x^{2} - 15 x + 9

\frac{- \frac{π}{3} + \frac{2 π}{3}}{2}

What are the solutions to the equation -3y^2=6y+2 ?
The solutions to the equation -3y^2=6y+2 are y=-(3+sqrt(3))/3 ,y=-(3-sqrt(3))/3
Find the zeros of -3y^2=6y+2
The zeros of -3y^2=6y+2 are y=-(3+sqrt(3))/3 ,y=-(3-sqrt(3))/3