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

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

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

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

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solve for

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

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

Solution steps

- x^{2} + 2 x = y

Move

y

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 1 ) ( - y )}{2 ( - 1 )}

Simplify

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

x_{1, 2} = \frac{- 2 \pm 2 1 - y}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 2 + 2 1 - y}{2 ( - 1 )}, x_{2} = \frac{- 2 - 2 1 - y}{2 ( - 1 )}

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

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

The solutions to the quadratic equation are:

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

x^{2} - 4 (2 + x) = 13

3 x^{2} - 10 x = 6 x - x^{2} - 15

x, (x + y) (x - y) = (x - y) 2

- x^{2} + x + 6 = 0

x^{2} + 20 x - 25 = 0

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