Quadratic equation within quadratic Edited version:- 
If both roots of equation $$4x^2 -20px+(25p^2 +15p- 66)=0 $$ are less than $2$, find the value of $p$. How to solve such type of equations?
 A: Hint : 
Rewrite as $(2x - 5p)^2 = 66 - 15p$, 
so that $2x - 5p = \pm \sqrt{66-15p}$.
Now you want $66 - 15p \gt 0$ and you want $x_1$ and $x_2$ to be less than 2.
I'm sure you can take it from here.
Another way would be to use the condition for both roots of a quadratic $ax^2 + bx + c = 0$ to be less than 2 are (1) Discriminant positive, (2) $ 2 \gt \frac{-b}{2a}$ and (3) $af(2) \gt 0$ where $f$ is the quadratic
A: The roots are $$\frac{1}{2} \left(5 p\pm\sqrt{3} \sqrt{22-5 p}\right)$$ where clearly the biggest root is $\frac{1}{2} \left(5 p+\sqrt{3} \sqrt{22-5 p}\right)$. Now if this root is smaller than $2$ the other root will also be smaller than $2$. 
\begin{align}
\frac{1}{2} \left(5 p+\sqrt{3} \sqrt{22-5 p}\right)&<2\\
5 p+\sqrt{3} \sqrt{22-5 p}&<4\\
\sqrt{3} \sqrt{22-5 p}&<4-5p\\
3 (22-5 p)&<(4-5p)^2\\
66-15 p&<16 - 40 p + 25 p^2\\
0&<-50 - 25 p + 25 p^2\\
0&<-2 -  p + p^2\\
0&<(p+1)(p-2)\\
\end{align}
If $p<-1$ or $p>2$ will the inequality hold ... (why?) Also note that $\sqrt{3} \sqrt{22-5 p}<4-5p$ requires $0<4-5p$ i.e. $p<\frac45$. Hence only $p<-1$ is acceptable. On the other hand we require $p<\frac{22}{5}$ hence $$p<-1.$$
