Explanation:
if I understand you correctly, then let's call the missing numbers l, r (for left and right).
and
-2 = l × r
-5 = l + r
l = -r - 5
therefore,
-2 = (-r - 5) × r = -r² - 5r
that gives us the quadratic equation
-r² - 5r + 2 = 0
for the general solutions
x = (-b ± sqrt(b² - 4ac))/(2a)
we have here
a = -1
b = -5
c = 2
and x = r, of course
r = (5 ± sqrt(25 - 4×-1×2))/(2×-1) =
= (5 ± sqrt(25 + 8))/-2 = (5 ± sqrt(33))/-2
r1 = (5 + sqrt(33))/-2 = -5.372281323...
r2 = (5 - sqrt(33))/-2 = 0.372281323...
as l = -r - 5, we see that r1 = l2, and r2 = l1.
but these are the only solutions.
there are no integer number solutions.