Intersection points of two curves

im doing calc 2 homework, so its kinda funny im asking a question that is just algebra/precalc. i had an integral problem to find the area between two functions, y=x^2-6x and y=-x^2+2x. i needed to find the bounds of the definite integral and when i set them equal to each other i found that the x^2 and -x^2 cancel out, and then i end up just getting x=0. then going off of that i got that the intersection points were (0,0) and (0,0). First of all this is obviously wrong, and so i graphed it with wolframalpha and saw the actual points where (0,0) and (4,-8). i got the right answer when i did the problem with the bounds 0 and 4, but i still am confused on how to find these intersection points and dont want to fumble like this on the next test. hopefully this isnt a really stupid question haha

really? I say that x² and -x² don't cancel out when equaling:

x²-6x=-x²+2x
2x²=8x
x²=4x
x(x-4)=0
x=0 , x=4

the integration is easy to do later
as you said it was ok.

...EXCEPT

x^2 - 6x = - x^2 + 2x

the (x^2) terms DO NOT CANCEL...positive on one side, negative on the other ? What are you thinking ?

2x^2 - 8x = 0

x^2 - 4x = 0

x(x - 4) = 0

x = 0 or x = 4

Can you find your limits of integration now ? I hope so !

id est

x^2 and -x^2 do not cancel out

x^2 - 6x = -x^2 + 2x
x^2 + x^2 - 6x - 2x = 0
2x^2 - 8x = 0
x^2 - 4x = 0
x * (x - 4) = 0
x = 0 , 4