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

The solutions to the equation solvefor y,x=2y-y^2 are y=1-sqrt(-x+1),y=sqrt(-x+1)+1

Find the zeros of solvefor y,x=2y-y^2

The zeros of solvefor y,x=2y-y^2 are y=1-sqrt(-x+1),y=sqrt(-x+1)+1

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solve for

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

y = 1 - - x + 1, y = - x + 1 + 1

Solution steps

x = 2 y - y^{2}

Switch sides

2 y - y^{2} = x

Move

x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

2^{2} - 4 (- 1) (- x) : 2 1 - x

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 1 - - x + 1, y = - x + 1 + 1

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

4 x^{2} - 4 x - 11 = - 8

4 x^{2} - 4 x + 2 = 0

x^{2} - 12 x - 160 = 0

x^{2} + 6 x + 1 = 8

What are the solutions to the equation solvefor y,x=2y-y^2 ?
The solutions to the equation solvefor y,x=2y-y^2 are y=1-sqrt(-x+1),y=sqrt(-x+1)+1
Find the zeros of solvefor y,x=2y-y^2
The zeros of solvefor y,x=2y-y^2 are y=1-sqrt(-x+1),y=sqrt(-x+1)+1