1. **State the problem:** Find the range of values of $t \in \mathbb{R}$ such that the point $(5,t)$ lies inside the circle defined by $$(x-3)^2 + (y+2)^2 = 29.$$\n\n2. **Recall the circle equation:** The circle has center $C = (3,-2)$ and radius $r = \sqrt{29}$. A point $(x,y)$ lies inside the circle if $$ (x-3)^2 + (y+2)^2 < 29.$$\n\n3. **Substitute the point $(5,t)$ into the inequality:**\n$$ (5-3)^2 + (t+2)^2 < 29.$$\n\n4. **Simplify:**\n$$ 2^2 + (t+2)^2 < 29,$$\n$$ 4 + (t+2)^2 < 29.$$\n\n5. **Isolate the squared term:**\n$$ (t+2)^2 < 29 - 4,$$\n$$ (t+2)^2 < 25.$$\n\n6. **Solve the inequality:**\nSince $(t+2)^2 < 25$, we take square roots carefully:\n$$ -5 < t+2 < 5.$$\n\n7. **Subtract 2 from all parts:**\n$$ -5 - 2 < t < 5 - 2,$$\n$$ -7 < t < 3.$$\n\n**Final answer:** The values of $t$ such that $(5,t)$ lies inside the circle are $$\boxed{-7 < t < 3}.$$