For vectors to be parallel they need to have the same slope.
vector i+j has slope: 1/1= 1
so slope of v has to be 1 which is:
(5t+6)/(3t^2-2t+8)=1
So it is a second degree polynomial equation to solve.
civodul
Find the values of t for which the velocity of the car is parallel to the vector (i + j)
v = (3r^2 - 2t + 8)i + (5t + 6)j ms^-1
Its doing my head in not being able to work this out