Consider the equation :
$$\mid x-5\mid^2+5\mid x-5\mid-24=0$$
The sum of all the real roots of the above equationis :

Let's consider x-5 as 'p'

Case 1: $$p \ge\ $$ 0

$$\mid x-5\mid^2+5\mid x-5\mid-24=0$$

$$p^2\ +5p\ -24\ =0\ $$

$$p^2\ +8p\ -3p\ -24\ =0\ $$

$$p\left(p+8\right)-3\left(p+8\right)\ =0\ $$

$$\left(p+8\right)\left(p-3\right)$$ = 0

p=-8 and p=3

x-5=3,x=8 This is a real root since x is greater than 5.

x-5=-8, x=-3. This root can be negated because x is not greater than 5.

Case 1: $$p < $$ 0

$$p^2\ -5p\ -24\ =0\ $$

$$p^2\ -8p\ +3p\ -24\ =0\ $$

p=8, -3

x-5=8, x=13. This root can be negated because x is not less than 5

x-5=-3, x=2. This is a real root because x is less than 5.

The sum of the real roots = 8+2=10

D is the correct answer.