What are the solutions to the equation 10y^2=1-3y ?

The solutions to the equation 10y^2=1-3y are y= 1/5 ,y=-1/2

Find the zeros of 10y^2=1-3y

The zeros of 10y^2=1-3y are y= 1/5 ,y=-1/2

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10 y^{2} = 1 - 3 y

y = \frac{1}{5}, y = - \frac{1}{2}

Decimal Notation

y = 0.2, y = - 0.5

Solution steps

10 y^{2} = 1 - 3 y

Move

3 y

to the left side

10 y^{2} + 3 y = 1

Move

1

to the left side

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

Solve with the quadratic formula

y_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 10 ( - 1 )}{2 \cdot 10}

3^{2} - 4 \cdot 10 (- 1) = 7

y_{1, 2} = \frac{- 3 \pm 7}{2 \cdot 10}

Separate the solutions

y_{1} = \frac{- 3 + 7}{2 \cdot 10}, y_{2} = \frac{- 3 - 7}{2 \cdot 10}

y = \frac{- 3 + 7}{2 \cdot 10} : \frac{1}{5}

y = \frac{- 3 - 7}{2 \cdot 10} : - \frac{1}{2}

The solutions to the quadratic equation are:

y = \frac{1}{5}, y = - \frac{1}{2}

y^{2} = 2 y + 3

3 x (x - 2) - (x + 3) = 23 (x - 3)

x^{2} = 21 - 4 x

(x - 1) (x + 12) = 42

- 3 y + 4 = y^{2}

What are the solutions to the equation 10y^2=1-3y ?
The solutions to the equation 10y^2=1-3y are y= 1/5 ,y=-1/2
Find the zeros of 10y^2=1-3y
The zeros of 10y^2=1-3y are y= 1/5 ,y=-1/2