so eqn < 0 just means negative curve graph and does not cut x axis

right..... if the y = ax^2 + b x + c and the question never say anything, then it could be any of the six cases... if a > 0 then it should be any of the U shapes. If a < 0 then it should be any of the n shapes

If clear, then we can discuss about the (m + 3)(m - 2).... we have different method of solving this type of inequality... What you teacher taught you?

my cher say d= eg; (m+3)(m-2)>0 it is (m+3)>0 , (m-2)>0 or (m+3)<0 ,(m-2)<0
(positive x positive or neg x neg)
then for (m+3)(m-2)<0,
m+3>0, m-2<0 or m+3<0 m-2>0
(pos x neg or neg x pos)
so for this qn
(m+3)(m-2)>0
so since its >0 i put
positive x positive >0
m+3>0 and m-2>0
m>-3 and m>2
or
negative x negative >0
m+3<0 and m-2<0
m<-3 and m<2

this method is very difficult to learn anyway, in this case is should be (m + 3)(m- 2) <0
1. There are two factors: (m + 3) and (m -2)
2. to have a negative product (< 0), one of them > 0 then the other one must be < 0.
3. two cases
4. case 1 (m + 3) > 0 and (m - 2) < 0
m > -3 and m < 2
which part of the number line satisfy both?
-3 < m < 2
5. case 2 (m + 2) < 0 and (m - 2) > 0
m < -3 and m > 2
which part of the number line satisfy both?
none....
6. final set of solution: -3 < m < 2