1. State the problem. We are asked to find the Laplace transform of $3e^{-2t}$. 2. Formula and rules. The Laplace transform is defined by $$\mathcal{L}\{f(t)\}(s)=\int_0^{\infty} e^{-st} f(t)\,dt$$ For an exponential $e^{at}$ the rule is $\mathcal{L}\{e^{at}\}(s)=\frac{1}{s-a}$ provided $\text{Re}(s)>\text{Re}(a)$. 3. Apply the definition. Compute $\mathcal{L}\{3e^{-2t}\}(s)=\int_0^{\infty} 3 e^{-st} e^{-2t}\,dt$. Combine the exponents to get $3\int_0^{\infty} e^{-(s+2)t}\,dt$. 4. Evaluate the integral. Use $\int_0^{\infty} e^{-\alpha t}\,dt=\frac{1}{\alpha}$ for $\text{Re}(\alpha)>0$. Here $\alpha=s+2$, so the integral equals $3\cdot \frac{1}{s+2}$. 5. Final answer. Thus $\mathcal{L}\{3e^{-2t}\}(s)=\frac{3}{s+2}$ and the region of convergence is $\text{Re}(s)>-2$.