Explanation:
I guess method 1 means to deal with whole factors.
x + 5 = (x - 2)(x + 5)
for (x + 5) <> 0 we can divide both sides by this factor :
1 = x - 2
x = 3
for the second solution we deal with
x + 5 = 0
x = -5
so, for x = -5 and x = 3 both functions deliver the same output, and these are the intersection points.
method 2 : we multiply the expression out and solve it then
x + 5 = (x - 2)(x + 5)
x + 5 = x² + 5x - 2x - 10 = x² + 3x - 10
0 = x² + 2x - 15
the general solution to such a square equation is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = 2
c = -15
x = (-2 ± sqrt(2² - 4×1×-15))/(2×1) =
= (-2 ± sqrt(4 + 60))/2 = (-2 ± sqrt(64))/2 = (-2 ± 8)/2 =
= -1 ± 4
x1 = -1 + 4 = 3
x2 = -1 - 4 = -5
and you get the 2 solutions again. as expected, they are the same as with method 1, of course.