What are the solutions to the equation solvefor x,y=(x-3)^2-4 ?

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

Find the zeros of solvefor x,y=(x-3)^2-4

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

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solve for

x, y = (x - 3)^{2} - 4

x = y + 4 + 3, x = - y + 4 + 3

Solution steps

y = (x - 3)^{2} - 4

Switch sides

(x - 3)^{2} - 4 = y

Move

4

to the right side

(x - 3)^{2} = y + 4

For

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

the solutions are

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

Solve

x - 3 = y + 4 : x = y + 4 + 3

Solve

x - 3 = - y + 4 : x = - y + 4 + 3

The solutions to the quadratic equation are:

x = y + 4 + 3, x = - y + 4 + 3

x, y^{4} + 7 x y^{2} = x^{2}

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

x^{2} - 3.5 x = 0

4 x^{2} - 12 x - 280 = 0

3 x^{2} + x - 4 = 0

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