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

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

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

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

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4 - y^{2} = 2 - y

y = - 1, y = 2

Solution steps

4 - y^{2} = 2 - y

Move

y

to the left side

4 - 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 ( - 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

3 x^{2} = 33 x

x^{2} - 5 x - 75 = 0

3 (x + 7)^{2} + 4 = 79

x, y = (x - 2)^{2}

x^{2} + 3 x = 2 (1 - x^{2})

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