What are solutions of the equation u(v(t))=0 if u(t)=t^2-16 and v(t)=t+2 ?

We'll substitute v(t) by it's
expression:

u(v(t)) =
u((t+2))

u((t+2)) = (t+2)^2 -
16

We'll notice that we have a difference of
squares:

u((t+2))
= (t+2-4)(t+2+4)

u((t+2))
= (t-2)(t+6)

We'll solve the
equation:

u((t+2)) = 0 <=> (t-2)(t+6) =
0

We'll set each factor as
zero:

t-2 = 0

t =
2

t+6=0

t=-6

The
solutions of the equation u(v(t)) = 0 are: {-6 ;
2}.