1. **State the problem:** Simplify the expression $$\sqrt{\left(6-\frac{3}{2}\right)2}+\sqrt{-t}$$. 2. **Simplify inside the first square root:** Calculate $$6-\frac{3}{2}$$. $$6=\frac{12}{2}$$ so $$6-\frac{3}{2}=\frac{12}{2}-\frac{3}{2}=\frac{12-3}{2}=\frac{9}{2}$$. 3. **Multiply by 2:** $$\left(6-\frac{3}{2}\right)2=\frac{9}{2} \times 2$$. Show cancellation: $$\frac{9}{\cancel{2}} \times \cancel{2}=9$$. 4. **Simplify the first square root:** $$\sqrt{9}=3$$. 5. **Analyze the second square root:** $$\sqrt{-t}$$ is real only if $$-t \geq 0$$, which means $$t \leq 0$$. 6. **Final simplified expression:** $$3 + \sqrt{-t}$$ with the condition $$t \leq 0$$ for real values. **Answer:** $$3 + \sqrt{-t}$$ where $$t \leq 0$$.