The value of p for which both the roots of the equation 4x2-20px+(25p2+15p-66)=0 are less than 2, lies in

Let f(x)=4x2-20px+(25p2+15p-66)=0  ...(i)The roots of (i) are real ifb2-4ac=400p2 -16(25p2+15p-66)=16(66-15p)≥0⇒p≤22/5               ...(ii)Both roots of (i) are less than 2. Therefore f(2)>0 and sum of roots <4⇒4.22-20 p.2+(25p2+15p-66)>0 and 20p4<4⇒p2-p-2>0 and p<4/5⇒(p+1)(p-2)>0 and p<4/5⇒p<-1 or p>2 and p<4/5⇒p<-1 ...(iii)From (ii) and (iii), we get p<-1 i.e. p∈(-∞, -1).