The sum of all the roots of the equation $$|x^2 - 8x + 15| - 2x + 7 = 0$$ is

Write the given equation in two parts:

$$|x^{2}-8x+15| - 2x + 7 = 0$$

The quadratic inside the absolute value factors as

$$x^{2}-8x+15 = (x-5)(x-3)$$

Its roots are $$x = 3$$ and $$x = 5$$, and the parabola opens upward, so

$$x^{2}-8x+15 \lt 0 \text{ for } 3 \lt x \lt 5,\qquad x^{2}-8x+15 \ge 0 \text{ for } x \le 3 \text{ or } x \ge 5$$

Therefore split the equation into two cases.

Case 1: $$x \le 3$$ or $$x \ge 5$$

Here $$|x^{2}-8x+15| = x^{2}-8x+15$$, so

$$x^{2}-8x+15 - 2x + 7 = 0 \\ \Rightarrow x^{2} - 10x + 22 = 0$$

Using the quadratic formula,

$$x = \frac{10 \pm \sqrt{100-88}}{2} = \frac{10 \pm 2\sqrt{3}}{2} = 5 \pm \sqrt{3}$$

Check which roots satisfy the domain of this case:

$$5+\sqrt{3} \approx 6.732 \ge 5 \quad \text{(acceptable)}$$

$$5-\sqrt{3} \approx 3.268 \quad (3 \lt 3.268 \lt 5,\; \text{reject})$$

Thus Case 1 contributes the root $$x = 5+\sqrt{3}$$.

Case 2: $$3 \lt x \lt 5$$

Now $$|x^{2}-8x+15| = -(x^{2}-8x+15)$$, so

$$-(x^{2}-8x+15) - 2x + 7 = 0 \\ \Rightarrow -x^{2} + 8x - 15 - 2x + 7 = 0 \\ \Rightarrow -x^{2} + 6x - 8 = 0 \\ \Rightarrow x^{2} - 6x + 8 = 0$$

Again apply the quadratic formula:

$$x = \frac{6 \pm \sqrt{36-32}}{2} = \frac{6 \pm 2}{2} = 3 \pm 1$$

So the possible solutions are $$x = 4$$ and $$x = 2$$. Only $$x = 4$$ lies in the interval $$3 \lt x \lt 5$$, hence it is the sole root from Case 2.

The complete set of roots is $$\{\,4,\;5+\sqrt{3}\,\}$$.

Sum of all roots:

$$4 + \bigl(5+\sqrt{3}\bigr) = 9 + \sqrt{3}$$

Therefore the required sum is $$9 + \sqrt{3}$$, which matches Option B.