$\newcommand\ss{\text{ss}}\newcommand\Z{\mathbb Z}$For real $t\ge0$, let $X_t:=mt+W_t$, where $m$ is a real number and $W_\cdot$ is a standard Brownian motion. So, $X_\cdot$ is a drifted Brownian motion starting at $0$ with the constant drift coefficient $m$. 

For real $c$, let $T_c:=\min\{t\ge0\colon X_t=c\}$. 

According to formula 3.0.2 on p. 309 in [Handbook of Brownian Motion - Facts and Formulae (Second Edition) by A. Borodin and P. Salminen][1], the pdf, say $p_{a,b}$, of the random variable (r.v.) $T_{a,b}:=\min(T_a,T_b)$ is given by the formula 
$$p_{a,b}(u)=e^{-m^2u/2}\big(e^{ma}\ss_u(b,b-a)+e^{mb}\ss_u(-a,b-a)\big)$$
for real $u\ge0$, where $-\infty<a<0<b<\infty$ and, according to the notation list on p. 641 of the mentioned handbook, 
$$\ss_u(c,d):=\sum_{k\in\Z}\frac{d-c+2k d}{u^{3/2}\sqrt{2\pi}}
\exp\Big\{-\frac{(d-c+2k d)^2}{2u}\Big\}.$$ 
So, for real $t>0$,
$$Q_{a,b}(t):=P(T_{a,b}>t)=\int_t^\infty du\,p_{a,b}(u). \tag{1}$$

Thus, the probability in question is 
$$P(T_a\le t,T_b\le t)=1+P(T_a>t,T_b>t)-P(T_a>t)-P(T_b>t) \\
=1+Q_{a,b}(t)-Q_{a,\infty-}(t)-Q_{-\infty+,b}(t),$$
with $Q_{a,b}(t)$ given by (1). 

Using e.g. formula 2.0.2 on p. 295 of the mentioned handbook, the expressions for $P(T_a>t)=Q_{a,\infty-}(t)$ and $P(T_b>t)=Q_{-\infty+,b}(t)$ can be simplified as follows: 
$$P(T_z>t)
=\frac1{2z}
\Big(1-e^{2 m z}-\text{erf}\Big(\frac{m t-z}{\sqrt{2t}}\Big)+e^{2 m z} \text{erf}\Big(\frac{m t+z}{\sqrt{2t}}\Big)\Big)$$
for $z\in\{a,b\}$. 



  [1]: https://www.semanticscholar.org/paper/Handbook-of-Brownian-Motion-Facts-and-Formulae-Borodin-Salminen/f05f3576e51a889e7ca02e98cec482f441538149