Find standard deviation $s$ such that $\mathbb{P}(X \in [-t, t]) = 1 - \alpha$ for $X \sim \mathcal{N}(0, s)$

Question

Let $t >0$ and $\alpha \in (0, 1)$.

I am looking for the standard deviation $s > 0$ such that $\mathbb{P}(X \in [-t, t]) = 1 - \alpha$ for any $X \sim \mathcal{N}(0, s)$.

Let $s$ be a candidate value.

$$ \mathbb{P}(X \in [-t, t]) = \int_{-t}^t \frac{1}{\sqrt{2\pi} \cdot s}\exp\left(- \frac{u^2}{2 s^2} \right)\textrm{d}u =: F(s) $$

So I have to find $s$ such that $F(s) = 1 - \alpha$.

Is it possible to do that without using gradient descent methods? What would be the intelligent way of doing this?

Thanks a lot.