1. We need to solve $|x^2-8x+15|-2x+7=0$ which is equivalent to $|x^2-8x+15|=2x-7.$

Since the left side is non-negative, we must have $2x-7\ge 0 \implies x\ge \frac{7}{2}.$

Also, $x^2-8x+15=(x-3)(x-5).$ So we split into cases depending on the sign of $(x-3)(x-5)$.

2. Case 1: $x^2-8x+15\ge 0$

This happens when $x\le 3 \quad \text{or} \quad x\ge 5.$ But we also need $x\ge \frac72$, so from this case only $x\ge 5$ is possible.

Then $|x^2-8x+15|=x^2-8x+15.$ So the equation becomes $x^2-8x+15-2x+7=0$ $x^2-10x+22=0.$ Solving, $x=\frac{10\pm\sqrt{100-88}}{2}=\frac{10\pm\sqrt{12}}{2}=5\pm\sqrt3.$ Now check validity for this case $x\ge 5$:

• $5+\sqrt3 \ge 5$ is valid.
• $5-\sqrt3 < 5$ is not valid.

So from Case 1, the valid root is $x=5+\sqrt3.$

3. Case 2: $x^2-8x+15<0$

This happens when $3<x<5.$ Since also $x\ge \frac72$, this reduces to $\frac72 \le x<5.$

Then $|x^2-8x+15|=-(x^2-8x+15)=-x^2+8x-15.$ So the equation becomes $-x^2+8x-15-2x+7=0$ $-x^2+6x-8=0$ $x^2-6x+8=0$ $(x-2)(x-4)=0.$ Thus, $x=2,4.$ Now check validity for this case $\frac72 \le x<5$:

• $x=2$ is invalid.
• $x=4$ is valid.

So from Case 2, the valid root is $x=4.$

4. Therefore, all roots are $x=4,\quad x=5+\sqrt3.$ Their sum is $4+(5+\sqrt3)=9+\sqrt3.$

Hence the correct option is $\boxed{B: 9+\sqrt3}.$