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

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

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

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

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solve for

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

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

Solution steps

y = 2 x - x^{2}

Switch sides

2 x - x^{2} = y

Move

y

to the left side

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

Write in the standard form

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

- 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} - 2 x - 2 = 5

- x^{2} + 4 x + 6

x^{2} - 2 x + 193 = 0

8 (x^{2} + 4) - 2 x - 2 = 2 x^{2} - 5 x + 30

x \cdot x + 4 = x + 2 \cdot x + 1

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