What are the solutions to the equation 14y^2=1-5y ?

The solutions to the equation 14y^2=1-5y are y= 1/7 ,y=-1/2

Find the zeros of 14y^2=1-5y

The zeros of 14y^2=1-5y are y= 1/7 ,y=-1/2

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14 y^{2} = 1 - 5 y

y = \frac{1}{7}, y = - \frac{1}{2}

Solution steps

14 y^{2} = 1 - 5 y

Move

5 y

to the left side

14 y^{2} + 5 y = 1

Move

1

to the left side

14 y^{2} + 5 y - 1 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 14 ( - 1 )}{2 \cdot 14}

5^{2} - 4 \cdot 14 (- 1) = 9

y_{1, 2} = \frac{- 5 \pm 9}{2 \cdot 14}

Separate the solutions

y_{1} = \frac{- 5 + 9}{2 \cdot 14}, y_{2} = \frac{- 5 - 9}{2 \cdot 14}

y = \frac{- 5 + 9}{2 \cdot 14} : \frac{1}{7}

y = \frac{- 5 - 9}{2 \cdot 14} : - \frac{1}{2}

The solutions to the quadratic equation are:

y = \frac{1}{7}, y = - \frac{1}{2}

4^{- 5}

x + 5 + 6 x^{- 1}

a^{4} + 5 a^{2} - 36

7 z + 15 = 32

\frac{1}{12}

What are the solutions to the equation 14y^2=1-5y ?
The solutions to the equation 14y^2=1-5y are y= 1/7 ,y=-1/2
Find the zeros of 14y^2=1-5y
The zeros of 14y^2=1-5y are y= 1/7 ,y=-1/2