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

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

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

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

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solve for

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

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

Solution steps

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

Switch sides

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

For

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

the solutions are

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

Solve

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

Solve

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

The solutions to the quadratic equation are:

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

9^{2} + b^{2} = 1 0^{2}

12 x^{2} - 4 x - 5 = 0

48 = x^{2} + 8 x

- 5 m^{2} = - 6 m

x^{2} - 7 x = 9

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