Proof that for $\alpha > 0$ and $\tau \geq 0$ , P($s$) = $(s+\alpha+ e^{-s\tau})$ has no solution less than zero

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P($s$) = $(s+\alpha+ e^{-s\tau})$

Proof that for $\alpha > 0$ and $\tau \geq 0$ the polynomial has no solution less than zero.

I am having difficulty proving that for $\alpha > 0$ and $\tau \geq 0$, the polynomial below doesn't lead to a real negative part (in case the solution was imaginary).

$s+\alpha+ e^{-s\tau} = 0$