I have reached two integrals:
$$\int_{(x+L)/(2c\sqrt{t})}^\infty e^{-z^2} dz + \int_{-\infty}^{(x-L)/(2c\sqrt{t})} e^{-z^2} dz$$
Now the first evaluates just fine to $\frac{\sqrt{\pi}}2\left(1-\operatorname{erf}\left(\frac{x+L}{2c\sqrt{t}}\right)\right)$, but I don't know how to evaluate the right integral.
Note: The integrals are confirmed correct.
Can I change of variable $z=-q$ and get $\int_{-(x-L)/(2c\sqrt{t})}^\infty e^{-q^2} dq$ perhaps? Since The $z$ is squared
Noting seeming to come out right? Probably because there is an error lurking around.
Using the identity, $$\int_{a}^{b} e^{-z^2} dz = \frac{\sqrt{\pi}}{2}(\operatorname{erf}(b) - \operatorname{erf}(a)).$$ If you can recognize what $a$ and $b$ are, this shouldn't be very difficult.
Now for the problem at hand, the integral $\int_{-\infty}^{(x-L)/(2c\sqrt{t})} e^{-z^2} dz$, like you said, is equivalent to, $$\int_{-(x-L)/(2c\sqrt{t})}^\infty e^{-q^2} dq = \frac{\sqrt{\pi}}{2}\operatorname{erfc} \left ( \frac{(x-L)}{2c\sqrt{t}} \right ) = \frac{\sqrt{\pi}}{2}\left ( 1 - \operatorname{erf}\left ( \frac{(x-L)}{2c\sqrt{t}} \right ) \right).$$