Find the points of intersection of the circles. x^2+y^2-18x-4y+35=0 and x^2+y^2+2x+6y-15=0?

Subtract the first equation from the second:

20x + 10y - 50 = 0
2x + y = 5 .... simplified
y = -2x + 5 .... slope-intercept form

Any (x,y) satisfying both quadratic equations must be on this line. Now you can substitute (-2x + 5) for each y in one of those to get a quadratic in x alone. Try the second since it has smaller coefficients:

x² + (-2x + 5)² + 2x + 6(-2x + 5) - 15 = 0
x² + 4x² - 20x + 25 + 2x - 12x + 30 - 15 = 0
5x² - 30x + 40 = 0
x² - 6x + 8 = 0
x² - 6x + 9 = 1 .... add 1 to complete the square on the left
(x - 3)² = 1
x = 3 ± 1
So x is 2 or 4, and y=-2x + 5 is 1 or -3

(2, 1) and (4, -3) are the points.

x² + y² - 18x - 4y + 35 = 0 ...... [1]
x² + y² + 2x + 6y - 15 = 0 ...... [2]

[2] - [1] :
20x + 10y - 50 = 0
2x + y - 5 = 0
y = 5 - 2x ...... [3]

Substitute [3] into [2] :
x² + (5 - 2x)² + 2x + 6(5 - 2x) - 15 = 0
x² + 25 - 20x + 4x² + 2x + 30 - 12x - 15 = 0
5x² - 30x + 40 = 0
x² - 6x + 8 = 0
(x - 2)(x - 4) = 0
x = 2 or x = 4

When x = 2, substitute it into [3] :
y = 5 - 2(2)
y = 1

When x = 4, substitute it into [3] :
y = 5 - 2(4)
y = -3

The points of intersection : (2, 1) and (4, -3)