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

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

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

The zeros of -y=2-y^2 are y=-1,y=2

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- y = 2 - y^{2}

y = - 1, y = 2

Solution steps

- y = 2 - y^{2}

Switch sides

2 - y^{2} = - y

Move

y

to the left side

2 - y^{2} + y = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = - 1, y = 2

9 - 2 x < 7 + 2 (x - 3)

\frac{b ^{- 2}}{b ^{4}}

(\frac{1}{y})^{2}

81 y^{4} - 256

2 x + 9 + 3 x + x = 5 x + 6

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