What is the answer to solvefor y,y-2sqrt(x^2-2x+2)+1=0 ?

The answer to solvefor y,y-2sqrt(x^2-2x+2)+1=0 is y=2sqrt(x^2-2x+2)-1

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solvefor y,y-2sqrt(x^2-2x+2)+1=0

solve for

y, y - 2 x^{2} - 2 x + 2 + 1 = 0

y = 2 x^{2} - 2 x + 2 - 1

Solution steps

y - 2 x^{2} - 2 x + 2 + 1 = 0

Move

2 x^{2} - 2 x + 2

to the right side

y + 1 = 2 x^{2} - 2 x + 2

Move

1

to the right side

y = 2 x^{2} - 2 x + 2 - 1

x + 5 + 3 x + 4 = 7

x = 12

10 = \frac{1}{2 x 1 - x ^{2}}

0.137 = Y^{\frac{1}{3}} \cdot 0.28

- (x + 1)^{\frac{4}{5}} + 2 = - 14

What is the answer to solvefor y,y-2sqrt(x^2-2x+2)+1=0 ?
The answer to solvefor y,y-2sqrt(x^2-2x+2)+1=0 is y=2sqrt(x^2-2x+2)-1