What are the solutions to the equation solvefor x,(x+y)^2=289 ?

The solutions to the equation solvefor x,(x+y)^2=289 are x=17-y,x=-17-y

Find the zeros of solvefor x,(x+y)^2=289

The zeros of solvefor x,(x+y)^2=289 are x=17-y,x=-17-y

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solve for

x, (x + y)^{2} = 289

x = 17 - y, x = - 17 - y

Solution steps

(x + y)^{2} = 289

For

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

the solutions are

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

Solve

x + y = 289 : x = 17 - y

Solve

x + y = - 289 : x = - 17 - y

The solutions to the quadratic equation are:

x = 17 - y, x = - 17 - y

10 - 5 x = 2 x^{2}

(y - 6)^{2} - 4 = 14

3 x^{2} - 27 = 0

(2 x + 3)^{2} = 6

u^{2} + u - 30 = 0

What are the solutions to the equation solvefor x,(x+y)^2=289 ?
The solutions to the equation solvefor x,(x+y)^2=289 are x=17-y,x=-17-y
Find the zeros of solvefor x,(x+y)^2=289
The zeros of solvefor x,(x+y)^2=289 are x=17-y,x=-17-y