1. Substitute to simplify

Let $t=e^x.$ Since $x\in\mathbb R$, we have $t=e^x>0.$

The given equation becomes $t^4+8t^3+13t^2-8t+1=0.$

So we need the positive roots of this quartic.

2. Factor the quartic

Assume $t^4+8t^3+13t^2-8t+1=(t^2+at-1)(t^2+bt-1).$

Expanding:

Compare coefficients with $t^4+8t^3+13t^2-8t+1.$

So,

Thus $a,b$ are roots of $u^2-8u+15=0,$ which gives $u=3,5.$

Hence, $t^4+8t^3+13t^2-8t+1=(t^2+3t-1)(t^2+5t-1).$

Therefore, $(e^{2x}+3e^x-1)(e^{2x}+5e^x-1)=0.$

3. Solve each quadratic in $t$

(i) $t^2+3t-1=0$

$t=\frac{-3\pm\sqrt{9+4}}{2}=\frac{-3\pm\sqrt{13}}{2}.$

Since $t>0$, only $t=\frac{-3+\sqrt{13}}{2}>0$ is valid.

So, $x=\ln\left(\frac{-3+\sqrt{13}}{2}\right).$

Now, $\frac{-3+\sqrt{13}}{2}<1$ (because $\sqrt{13}<5$), hence this $x<0$.

(ii) $t^2+5t-1=0$

$t=\frac{-5\pm\sqrt{25+4}}{2}=\frac{-5\pm\sqrt{29}}{2}.$

Since $t>0$, only $t=\frac{-5+\sqrt{29}}{2}>0$ is valid.

So, $x=\ln\left(\frac{-5+\sqrt{29}}{2}\right).$

Also, $\frac{-5+\sqrt{29}}{2}<1$ (because $\sqrt{29}<7$), hence this $x<0$.

4. Count the real solutions

We get exactly two valid positive values of $t=e^x$, hence exactly two real values of $x$.

Both satisfy $0<t<1$, so both corresponding $x=\ln t$ are negative.

5. Conclusion

The equation has two real solutions and both are negative.

Therefore, the correct option is: $\boxed{\text{A}}$

6. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.