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

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

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

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

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solve for

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

x = \frac{y - 4}{2} + 2, x = - \frac{y - 4}{2} + 2

Solution steps

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

Switch sides

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

Move

4

to the right side

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

Divide both sides by

2

(x - 2)^{2} = \frac{y - 4}{2}

For

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

the solutions are

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

Solve

x - 2 = \frac{y - 4}{2} : x = \frac{y - 4}{2} + 2

Solve

x - 2 = - \frac{y - 4}{2} : x = - \frac{y - 4}{2} + 2

The solutions to the quadratic equation are:

x = \frac{y - 4}{2} + 2, x = - \frac{y - 4}{2} + 2

x^{(2)} + 3 x + 2 = 0

6 c^{2} - 2 c - 3 = 1

x = x^{2} - 3

42 (p^{2} - 1) = 13 p

x (x + 3) = 180

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