A journey towards infinity

Solving the quadratic, $x^2 - (a-3)x + a = 0$ $\implies x = \dfrac {a-3 \pm \sqrt{(a-3)^2 -4a}}2$ . For the roots to be real,

$\begin{aligned} \implies (a-3)^2 - 4a & \ge 0 \\ a^2 - 10a + 9 & \ge 0 \\ (a-1)(a-9) & \ge 0 \\ \implies a \le 4 & \cup a \ge 9 \end{aligned}$

For both roots to be $> 2$ ,

$\begin{aligned} \frac {a-3 - \sqrt{(a-3)^2 -4a}}2 & > 2 \\ a-3 - \sqrt{a^2-10a+9} & > 4 \\ a-7 & > \sqrt{a^2-10a+9} & \small \color{#3D99F6} \text{Squaring both sides} \\ a^2 - 14a + 40 & > a^2-10a+9 \\ \implies a < 10 \end{aligned}$

Therefore, for both roots to be $>2$ , $a \in \boxed{[9, 10)}$ .