jmc

First, we evaluate the first pair of parentheses:

$$\begin{array}{rl}(t)\bowtie (t+2)& =(t+2)((t+2)-3)\\ & =(t+2)(t-1)\\ & =t^2+2t-t-2\\ & =t^2+t-2.\end{array}$$Next, we evaluate the second pair of parentheses:

$$\begin{array}{rl}(t+1)\bowtie (t+1)& =((t+1)+2)((t+1)-3)\\ & =(t+3)(t-2)\\ & =t^2+3t-2t-6\\ & =t^2+t-6.\end{array}$$Subtracting the two expressions and noting that certain terms cancel, we obtain $(t^2+t-2)-(t^2+t-6)=-2-(-6)=\boxed{4}$.