Find three consecutive numbers such that the sum of one-fourth the first and one-fifth the second is five less than one-seventh the third.

-10, -11 and -12
-12, -13 and -14
-14, -15 and -16

Consecutive are 1 apart

they are
x,x+1,x+2

(x)/4+(x+1)/5=-5+(x+2)/7
solve for x
multiply both sides by (4*5*7) or 140
35x+28(x+1)=-700+20(x+2)
expand
35x+28x+28=-700+20x+40
63x+28=20x-660
minus 20x both sides
43x+28=-660
minus 28 both sides
43x=-688
divide both sides by 43
x=-16
x+1=-15
x+2=-14

the numbers are
-16,-15,-14 (in order of first second third)