The parametric equations are:
x=t^2+3t-8, dx/dt=2t+3, and
The point (2,-1) is (x,y) so:
2=t^2+3t-8 and -1=2t^2-2t-5,
t^2+3t-10=0=(t+5)(t-2), 2t^2-2t-4=0=2(t^2-t-2)=2(t-2)(t+1). The common factor is t-2=0, so t=2.
This is the value of t which produces the coordinates (x,y)=(2,-1) and defines the tangent at (2,-1), so we substitute t=2 in dy/dx=6/7, the slope of the tangent.