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

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

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

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

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solve for

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

x = - y + 4 - 2, x = - - y + 4 - 2

Solution steps

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

Switch sides

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

Move

4

to the right side

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

Divide both sides by

- 1

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

For

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

the solutions are

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

Solve

x + 2 = - y + 4 : x = - y + 4 - 2

Solve

x + 2 = - - y + 4 : x = - - y + 4 - 2

The solutions to the quadratic equation are:

x = - y + 4 - 2, x = - - y + 4 - 2

x^{2} - 9 x - 20 = - 2 x - 2

(x - 36) (x + 36) = 0

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

v^{2} + 3 v - 2 = 0

x^{2} - 13 x + 24 = - x + 4

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