Express x in terms of y where sqrt(x-y)=3

Given the equation
$$\sqrt{x - y} = 3$$
Because equation degree is equal to = 1/2 - does not contain even numbers in the numerator, then
the equation has single real root.
We raise the equation sides to 2-th degree:
We get:
$$\left(\sqrt{x - y}\right)^{2} = 3^{2}$$
or
$$x - y = 9$$
Looking for similar summands in the left part:

x - y = 9

Move the summands with the other variables
from left part to right part, we given:
$$x = y + 9$$
We get the answer: x = 9 + y

All other 1 root(s) is the complex numbers.
do replacement:
$$z = x - y$$
then the equation will be the:
$$\sqrt{z} = 3$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$\sqrt{r e^{i p}} = 3$$
where
$$r = 9$$
- the magnitude of the complex number
Substitute r:
$$e^{\frac{i p}{2}} = 1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(\frac{p}{2} \right)} + \cos{\left(\frac{p}{2} \right)} = 1$$
so
$$\cos{\left(\frac{p}{2} \right)} = 1$$
and
$$\sin{\left(\frac{p}{2} \right)} = 0$$
then
$$p = 4 \pi N$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for z
Consequently, the solution will be for z:
$$z_{1} = 9$$
do backward replacement
$$z = x - y$$
$$x = y + z$$

The final answer:
$$x_{1} = y + 9$$