= Solution
The quartic is irreducible by <Eisenstein criterion> at two, so its <Galois group> is transitive on its four roots. Its <discriminant> is $1616=16\cdot101$, so five and seven are unramified primes for the <polynomial>. The squarefree factorizations are
$$
f(t)\equiv(t-1)(t^3+t^2+t-2)\pmod5,
\qquad
f(t)\equiv(t+2)(t+3)(t^2+2t-2)\pmod7.
$$
The cubic modulo five has no root in $\mathbb F_5$, hence is irreducible. The quadratic modulo seven has <discriminant> five, a nonsquare there. The <Frobenius cycle type> theorem therefore supplies a three-cycle and a transposition in the <group>.
Transitivity implies that its order is divisible by four, and the three-cycle implies divisibility by three. Its order divides $24$, so it is $12$ or $24$. The only subgroup of index two in $S_4$ is $A_4$: an index-two subgroup is the <kernel of a group homomorphism> of a nontrivial homomorphism to $C_2$, and all transpositions are conjugate and generate $S_4$, forcing this homomorphism to be the sign. The exhibited transposition excludes $A_4$. Hence
$$
\boxed{\operatorname{Gal}(t^4+2t+2/\mathbb Q)\cong S_4}.
$$
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