x²+15x+36<0
at first solve quadratic equation
D=b²-4ac= 225-4*1*36= 81
x=(-b+/-√D)/2a
x=(-15+/-√81)/2= (-15+/-9)/2
x1=(-15-9)/2=-12
x2=(-15+9)/2=-3
we can write x²+15x+36<0 as (x+12)(x+3)<0
(x+12)(x+3)<0 can be 2 cases, because for product to be negative one factor should be negative , and second factor should be positive
1 case) x+12<0, and x+3>0,
x<-12, and x>-3
(-∞, -12) and(-3,∞) gives empty set
or second case) x+12>0 and x+3<0
x>-12 and x<-3
(-12,∞) and (-∞,-3) they are crossing , so (-12, -3) is a solution of this inequality