What are the solutions to the equation 2y-3y^2+1=0 ?

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

Find the zeros of 2y-3y^2+1=0

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

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2 y - 3 y^{2} + 1 = 0

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

Decimal Notation

y = - 0.33333 \dots, y = 1

Solution steps

2 y - 3 y^{2} + 1 = 0

Write in the standard form

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

- 3 y^{2} + 2 y + 1 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

y_{1} = \frac{- 2 + 4}{2 ( - 3 )}, y_{2} = \frac{- 2 - 4}{2 ( - 3 )}

y = \frac{- 2 + 4}{2 ( - 3 )} : - \frac{1}{3}

y = \frac{- 2 - 4}{2 ( - 3 )} : 1

The solutions to the quadratic equation are:

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

x^{2} + 2 x + 17

x, 4 x^{2} - 2 x y^{3} + 3 y^{2} - 6 x + 5 = 0

x \cdot 2 x = 72

x^{2} + 2 x - 39

2 x - 1 - x + 2 = x^{2} - 8 - 2 x - 1

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