<s>[INST] 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?
 [/INST] 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 $$p<-1.$$
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