This looks like a problem derived from Bezier curves.
I assume that $x$, $a$, $b$, $c$, $d$ are all known, and we're trying to find $t$.
So, you just multiply out all the powers of $(1-t)$ and gather together the terms containing the same power of $t$. You will end up with four terms: one each for $t^3$, $t^2$, and $t$, and a constant term. You actually get:
(-a+3b-3c+d)t^3 + (3a-6b+3c)t^2 + (-3a+3b)t + a - x =0
In other words, you will end up with something of the form:
$$pt^3 + qt^2 + rt + s = 0$$
where $p=-a+3b-3c+d$, $q = 3a-6b+3c$, $r = -3a+3b$, and $s=a-x$.
This is a so-called "cubic" equation that you need to solve for $t$.
There are formulas that give you the solutions of cubic equations, just as there are formulas for solving quadratics, but the cubic ones are much more complicated. Some techniques are described on this Wikipedia page. Also, you can find many "cubic equation calculators" on the internet that will give you the solutions for specific given values of $x$, $a$, $b$, $c$, $d$.