(Calculus) What is the rectangular equation (by eliminating the parameter) of x=(t^2)+t and y=(t^2)-t?

Question

Question: Write the corresponding rectangular equation by eliminating the parameter:

x=(t^2) + t
y=(t^2) - t

Just can't seem to figure this one out :(

Answer 1

Subtracting you get: x-y = 2t
So t = (x-y)/2
Now plug this back into the first equation:
x = (x-y)²/4 + (x-y)/2
or
(x-y)² = 2(x+y)

Note:
By writing this as
y² - (2x+2)y + (x²-2x) = 0
You could also have:
y = (2x+1) +/- √(4x+1) where x >= -1/4

Addendum:
The understanding of that +/- is that for every x value, there are two t values that give that x, hence two values for y. For example, suppose that x = 2. Then the formula above gives 12 or 2. From the original problem, when x=2, then t = -3 or 2, since t²+t =2, and the two y=t²-t values that correspond to these are exactly 12 and 2.

Answer 2

x=(t^2) + t
y=(t^2) - t
t^2+t=x
t^2+t+1= x+1
(t+1)^2=x+1
t+1 =+/-sqrt(x+1)
t = -1 +/- sqrt(x+1)
y = (-1+/- sqrt(x+1))^2 -(-1+/- sqrt(x+1)
y = 1 +/- 2sqrt(x+1) +x+1+1 +/-sqrt(x+1)
y = 3+x +/-3sqrt(x+1)