What are the solutions to the equation solvefor x,y=(x-3)^2+1 ?

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

Find the zeros of solvefor x,y=(x-3)^2+1

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

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solve for

x, y = (x - 3)^{2} + 1

x = y - 1 + 3, x = - y - 1 + 3

Solution steps

y = (x - 3)^{2} + 1

Switch sides

(x - 3)^{2} + 1 = y

Move

1

to the right side

(x - 3)^{2} = y - 1

For

(g (x))^{2} = f (a)

the solutions are

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

Solve

x - 3 = y - 1 : x = y - 1 + 3

Solve

x - 3 = - y - 1 : x = - y - 1 + 3

The solutions to the quadratic equation are:

x = y - 1 + 3, x = - y - 1 + 3

- 0.5 x^{2} + 620 = 30 x + 270

4 x^{2} + 14 x - 3 = 3 x

8 x^{2} + 9 x - 4 = 0

20 x + 100 = 400 - x^{2}

- 4 x^{2} + 8 x + 12 = 0

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