Question

For what value(s) of t does the equality ⟨t^3−12t,0.25t^2+4⟩=⟨0,7⟩ hold true?
t=

Answer 1

The notation <....,....> usually represents vectors, so in this case
t^3 - 12t = 0 and .25t^2 + 4 = 7

from the first:
t(t^2 - 12) = 0
t = 0 or t = ±2√3

checking in the 2nd:
if t = 0
LS = 0 ≠ RS
if t = ± 2√3
LS = .25(12) + 4 = 7 = RS

so t = ± 2√3