The equation $\mathrm{e}^{4 x}+8 \mathrm{e}^{3 x}+13 \mathrm{e}^{2 x}-8 \mathrm{e}^{x}+1=0, x \in \mathbb{R}$ has :

Question

two solutions and both are negative

two solutions and only one of them is negative

four solutions two of which are negative

no solution

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