Without solving the following quadratic equation, find the greater value of $p$ for which the given equation has real and equal roots. $x^2+(p-3)x+p=0$.

Given equation $x^2+(p-x)x+p=0$

$\because$ Roots are real and equal, then $b^2-4ac=0$

$\therefore (p-3)^2-4\times 1\times p=0$

$\Rightarrow p^2+9-6p-4p=0$

$\Rightarrow p^2+9-10p=0$

$\Rightarrow p^2-10p+9=0$

$\Rightarrow p^2-9p-p+9=0$

$\Rightarrow p(p-9)-1(p-9)=0$

$\Rightarrow (p-9)(p-1)=0$

Hence, $p=9$ or $1$.