What are the solutions to the equation (x+n)^2+4=n^2 ?

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

Find the zeros of (x+n)^2+4=n^2

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

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(x + n)^{2} + 4 = n^{2}

x = n^{2} - 4 - n, x = - n^{2} - 4 - n

Solution steps

(x + n)^{2} + 4 = n^{2}

Move

4

to the right side

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

For

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

the solutions are

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

Solve

x + n = n^{2} - 4 : x = n^{2} - 4 - n

Solve

x + n = - n^{2} - 4 : x = - n^{2} - 4 - n

The solutions to the quadratic equation are:

x = n^{2} - 4 - n, x = - n^{2} - 4 - n

\frac{x ^{2}}{15} - \frac{2 x - 5}{3} = \frac{4}{5}

4 t^{2} - 11 t + 4 = 0

7 x^{2} + 4 x = 0

5 + 3 c^{2} = 17

24 π r - \frac{9 π}{2} r^{2} = 0

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