\( \left(t^{2}+4\right)^{2}-3\left(t^{2}+4\right)-40=0 \) The solution set is \( \{\tilde{\square}\} \).

We start with the equation   (t² + 4)² – 3(t² + 4) – 40 = 0. Notice that the expression t² + 4 appears twice. So, we make the substitution   u = t² + 4. The equation becomes   u² – 3u – 40 = 0. This is a quadratic equation in u. To solve it, we factor or use the quadratic formula. Factoring gives:   (u – 8)(u + 5) = 0. So, we have   u = 8  or  u = –5. Now, substitute back for u: 1. If u = 8:   t² + 4 = 8  ⟹  t² = 8 – 4 = 4  ⟹  t = 2 or t = –2. 2. If u = –5:   t² + 4 = –5  ⟹  t² = –5 – 4 = –9. Since t² = –9 has no real solutions (the solutions t = 3i and t = –3i are complex), if we are only considering real numbers we ignore this case. Thus, for real numbers the solution set is:   { –2, 2 }. If complex solutions are allowed, then we would also include t = 3i and t = –3i. However, typically such problems assume a real domain unless stated otherwise. Therefore, the solution set is { –2, 2 }.