What are the solutions to the equation solvefor x,x^2+(y-1)^2=4 ?

The solutions to the equation solvefor x,x^2+(y-1)^2=4 are x=sqrt(-y^2+2y+3),x=-sqrt(-y^2+2y+3)

Find the zeros of solvefor x,x^2+(y-1)^2=4

The zeros of solvefor x,x^2+(y-1)^2=4 are x=sqrt(-y^2+2y+3),x=-sqrt(-y^2+2y+3)

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solve for

x, x^{2} + (y - 1)^{2} = 4

x = - y^{2} + 2 y + 3, x = - - y^{2} + 2 y + 3

Solution steps

x^{2} + (y - 1)^{2} = 4

Move

(y - 1)^{2}

to the right side

x^{2} = 4 - (y - 1)^{2}

Simplify

4 - (y - 1)^{2} : - y^{2} + 2 y + 3

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = - y^{2} + 2 y + 3, x = - - y^{2} + 2 y + 3

x, x^{2} + 4 z^{2} - 8 y = 0

15 = - 0.08 x^{2} + 1.6 x + 10

d^{2} - 8 d + 9

10 = - 5 x^{2} + 2 x + 25

- 5 x - 5 = x^{2}

What are the solutions to the equation solvefor x,x^2+(y-1)^2=4 ?
The solutions to the equation solvefor x,x^2+(y-1)^2=4 are x=sqrt(-y^2+2y+3),x=-sqrt(-y^2+2y+3)
Find the zeros of solvefor x,x^2+(y-1)^2=4
The zeros of solvefor x,x^2+(y-1)^2=4 are x=sqrt(-y^2+2y+3),x=-sqrt(-y^2+2y+3)