A curve is given by:
$$x = 2t + 3 $$
$$y = t^3 - 4t$$
The point $A$ has parameter $t = -1$.
Line l is a tangent to the curve at $A$.
Line l cuts the curve at point $B$.
Find the value of $t$ at $B$.
I've worked out that the equation for l is: $2y + x = 7$ (which is correct). To work out the value of t at B, I rearranged the parametric equations to eliminate t and make y the subject. I did the same for the equation of line l (made y the subject). I equated the two together and formed an equation of a cubic curve: $x^3 - 9x^2 + 15x - 7 = 0$ (which is correct). The problem is that I do not know (or am expected to know) a method to find the roots of a cubic curve. I am wondering whether this is the best/correct approach, and whether there is an easier/alternative route to finding the answer.
We wish to find where our parameterized curve intersects the line $2y+x=7$. Let's parameterize that line by $x=2t+3,$ so we'll need $y=-t+2$. Under this parameterization, the $x$-values of our line and curve will be the same at any $t$, so we need only determine where the $y$-values are the same--i.e. find $t$ for which $$t^3-4t=-t+2\\t^3-3t-2=0.$$ Recall, though, that we already know they meet where $t=-1$, so we know we have a factor of $t+1$ in our cubic (that's why we went with this method, instead of yours). In particular, $$t^3-3t-2=(t+1)(t^2-t-2)=(t+1)^2(t-2),$$ so the cutting point occurs when $t=2$.