What are the solutions to the equation 5y^2-4=-y ?

The solutions to the equation 5y^2-4=-y are y= 4/5 ,y=-1

Find the zeros of 5y^2-4=-y

The zeros of 5y^2-4=-y are y= 4/5 ,y=-1

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5 y^{2} - 4 = - y

y = \frac{4}{5}, y = - 1

Solution steps

5 y^{2} - 4 = - y

Move

y

to the left side

5 y^{2} - 4 + y = 0

Write in the standard form

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

5 y^{2} + y - 4 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

y = \frac{- 1 + 9}{2 \cdot 5} : \frac{4}{5}

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

The solutions to the quadratic equation are:

y = \frac{4}{5}, y = - 1

\frac{y ^{6}}{y ^{4}}

- 2 < x \leq 1

5 ∣ 8 - x ∣ + 7 \leq 42

∣ x - 2 ∣ < 1

x^{2} + 6 x = 0

What are the solutions to the equation 5y^2-4=-y ?
The solutions to the equation 5y^2-4=-y are y= 4/5 ,y=-1
Find the zeros of 5y^2-4=-y
The zeros of 5y^2-4=-y are y= 4/5 ,y=-1