7. For every pair of real numbers $(x, y)$, the function $f(t)$ satisfies $f(x+y)=f(x)+f(y)+xy+1$. If $f(-2)$ $=-2$, then all integers $t$ that satisfy $f(t)=t$ are $t=$ $\qquad$ .
7. 1 or -2 Detailed Explanation: Let $x=y=0$ to get $f(0)=-1$; let $x=y=-1$, from $f(-2)=-2$ we get $f(-1)=-2$, and let $x=1, y=-1$ to get $f(1)=1$. Then let $x=1$, we have $f(y+1)=f(y)+y+2$ (1), so $f(y+1)-f(y)=y+2$, which means when $y$ is a positive integer, $f(y+1)-f(y)>0$. From $f(1)=1$, we know that for all positive integers $y$, $f(y)>0$, therefore for $y \in \mathbf{N}^{*}$, $f(y+1)=f(y)+y+2>y+1$, meaning for all positive integers $t$ greater than 1, $f(t)>t$. From (1), we get $f(-3)=-1$, $f(-4)=1$.

Next, we prove: when the integer $t \leqslant-4$, $f(t)>0$. Since $t \leqslant-4$, hence $-(t+2)>0$. From (1), we get: $f(t)-f(t+1)=$ $-(t+2)>0$,
i.e., $f(-5)-f(-4)>0, f(-6)-f(-5)>0, \cdots, f(t+1)-f(t+2)>0, f(t)-f(t+1)>0$, adding them up we get: $f(t)-f(-4)>0$. Since $t \leqslant-4$, thus $f(t)>t$. In summary: the integers $t$ that satisfy $f(t)=t$ are only $t=1$ or $t=-2$.