Find all values of p , so that 6 lies between roots of the equation x² + 2(p-3)x + 9 = 0

Dear Student,

x² + 2(p-3)x + 9 = 0
Let m & n be two roots of the equation

Comparing given equation with ax^​2 + bx + c =0

We get, a = 1 , b = 2(p-3) , c = 9

Now,
a) D > 0
b² - 4ac > 0
[2(p-3)]² - 4(1)(9) > 0
4(p² - 6p + 9) - 36 > 0
4p² - 24p + 36 - 36 > 0
p² - 6p > 0
p (p - 6) > 0
p $\in$ (-$\infty$ , 0) $\cup$ (6 , $\infty$)                ....................... (1) 

b) a f(6) < 0

1 . [6² + 2(p-3)6 + 9] < 0
36 + 12p - 36 + 9 < 0
4p + 3 < 0
p < - $\frac{3}{4}$
p $\in$( - $\infty$ , - $\frac{3}{4}$)                  .................................. (2)
c) Let m and n be the two roots of the equation
 sum of roots =  -2(p-3)

As per given condition: m < 6                        ...... (1)
                                      n > 6  $\Rightarrow$⇒-n < -6       ......... (2) 
Subtracting (2) from (1) we get,

m + n < 12
-2(p-3) < 12
p-3 > 6

p > 9
p $\in$(9 , $\infty$)                ........................... (3)

hence values of p satisfying (1) , (2) and (3) is p $\in$( - $\infty$ , - $\frac{3}{4}$)​ $\cup$ (9 , $\infty$)   ​

Regards