1. Given equation

$x + 1 - 2\log_2(3+2^x) + 2\log_4(10-2^{-x})=0$

We need the sum of all roots.

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2. Convert everything to base 2

Recall:

$\log_4 a = \frac{\log_2 a}{\log_2 4} = \frac{1}{2}\log_2 a$

So,

$2\log_4(10-2^{-x}) = 2\cdot \frac{1}{2}\log_2(10-2^{-x}) = \log_2(10-2^{-x})$

Hence the equation becomes

$x+1-2\log_2(3+2^x)+\log_2(10-2^{-x})=0$

Also,

$x = \log_2(2^x), \qquad 1=\log_2 2$

Thus

$x+1 = \log_2(2^x)+\log_2 2 = \log_2(2^{x+1})$

Therefore,

$\log_2(2^{x+1}) + \log_2(10-2^{-x}) - 2\log_2(3+2^x)=0$

Using log laws,

$\log_2\left(\frac{2^{x+1}(10-2^{-x})}{(3+2^x)^2}\right)=0$

So,

$\frac{2^{x+1}(10-2^{-x})}{(3+2^x)^2}=1$

That is,

$2^{x+1}(10-2^{-x})=(3+2^x)^2$

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3. Substitute $t=2^x$

Since $2^x>0$, let

$t=2^x \quad \Rightarrow \quad 2^{-x}=\frac{1}{t}$

Then the equation becomes

$2t\left(10-\frac{1}{t}\right)=(3+t)^2$

Simplify:

$20t-2=t^2+6t+9$

$t^2-14t+11=0$

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4. Solve the quadratic

$t^2-14t+11=0$

Using the quadratic formula,

$t=\frac{14\pm \sqrt{196-44}}{2} = \frac{14\pm \sqrt{152}}{2}=7\pm \sqrt{38}$

So the two possible values are

$t_1=7+\sqrt{38}, \qquad t_2=7-\sqrt{38}$

Both are positive, so both are valid since $t=2^x>0$.

Thus roots are

$x_1=\log_2(7+\sqrt{38}), \qquad x_2=\log_2(7-\sqrt{38})$

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5. Find the sum of roots

$x_1+x_2 = \log_2(7+\sqrt{38})+\log_2(7-\sqrt{38})$

$=\log_2\left((7+\sqrt{38})(7-\sqrt{38})\right)$

$=\log_2(49-38)=\log_2 11$

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6. Match with options

$\boxed{\log_2 11}$

So the correct option is B.

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7. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.