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

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

Find the zeros of y^2=2-y

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

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y^{2} = 2 - y

y = 1, y = - 2

Solution steps

y^{2} = 2 - y

Move

y

to the left side

y^{2} + y = 2

Move

2

to the left side

y^{2} + y - 2 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 1, y = - 2

- x^{2} - x + 30 = 0

16 = a^{2}

(x^{2} + x - 2) = 0

p^{2} + 3 = 0

16 x^{2} - 48 x + 36 = 0

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