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Let the roots of the quadratic be $x_1$ and $x_2$, with $x_1 < 2 < x_2$.

The quadratic equation is $x^2 - (5-a)x + a^2 - 11a - 46 = 0$.

By Vieta's formulas:
- $x_1 + x_2 = 5 - a$
- $x_1 x_2 = a^2 - 11a - 46$

Since the equation must have two real solutions, the discriminant must be non-negative:
$$\text{D = [-(5-a)] 2 - 4(a 2 - 11a - 46) = (5-a) 2 - 4(a 2 - 11a - 46)}$$
Expand:
$$(5-a)^2 = 25 - 10a + a^2$$
So,
$$D = 25 - 10a + a^2 - 4(a^2 - 11a - 46)$$
$$= 25 - 10a + a^2 - 4a^2 + 44a + 184$$
$$= (25 + 184) + (-10a + 44a) + (a^2 - 4a^2)$$
$$= 209 + 34a - 3a^2$$
So, $D \geq 0$:
$$209 + 34a - 3a^2 \geq 0$$

Now, the roots must be on opposite sides of $2$.
For a quadratic $ax^2 + bx + c = 0$, the value at $x = 2$ is:
$$f(2) = (2)^2 - (5-a)(2) + a^2 - 11a - 46$$
$$= 4 - 2(5-a) + a^2 - 11a - 46$$
$$= 4 - 10 + 2a + a^2 - 11a - 46$$
$$= (4 - 10 - 46) + (2a - 11a) + a^2$$
$$= (-52) - 9a + a^2$$
$$= a^2 - 9a - 52$$

For the roots to be on opposite sides of $2$, $f(2)$ must be negative (since the parabola opens upwards):
$$f(2) < 0 \implies a^2 - 9a - 52 < 0$$

So, we have two inequalities:
1. $209 + 34a - 3a^2 \geq 0$
2. $a^2 - 9a - 52 < 0$

Let's solve each.

First inequality:
$$209 + 34a - 3a^2 \geq 0$$
$$-3a^2 + 34a + 209 \geq 0$$
$$3a^2 - 34a - 209 \leq 0$$
Find the roots:
$$a = \frac{34 \pm \sqrt{(-34)^2 - 4 \cdot 3 \cdot (-209)}}{2 \cdot 3}$$
$$= \frac{34 \pm \sqrt{1156 + 2508}}{6}$$
$$= \frac{34 \pm \sqrt{3664}}{6}$$
$$\sqrt{3664} \approx 60.5$$
So,
$$a_1 = \frac{34 + 60.5}{6} \approx \frac{94.5}{6} \approx 15.75$$
$$a_2 = \frac{34 - 60.5}{6} \approx \frac{-26.5}{6} \approx -4.42$$
So, $3a^2 - 34a - 209 \leq 0$ for $-4.42 \leq a \leq 15.75$.

Second inequality:
$$a^2 - 9a - 52 < 0$$
Find the roots:
$$a = \frac{9 \pm \sqrt{81 + 208}}{2}$$
$$= \frac{9 \pm \sqrt{289}}{2}$$
$$= \frac{9 \pm 17}{2}$$
So,
$$a_1 = \frac{9 + 17}{2} = 13$$
a_2 = $\text{9}$ - 17}{2} = -4
$$$$
So, $a^2 - 9a - 52 < 0$ for $-4 < a < 13$.

The intersection of the two intervals:
$-4.42 \leq a \leq 15.75$ and $-4 < a < 13$

So, $a$ must satisfy $-4 < a < 13$.

Final answer:
All real numbers $a$ such that $-4 < a < 13$.