A quadratic equation has two distinct real solutions exactly when its discriminant is positive. For the quadratic equation $x^{2}-(r+7) x+r+87=0$, the discriminant is $\Delta=(r+7)^{2}-4(1)(r+87)=r^{2}+14 r+49-4 r-348=r^{2}+10 r-299$. Since $\Delta=r^{2}+10 r-299=(r+23)(r-13)$ which has roots $r=-23$ and $r=13$, then $\Delta>0$ exactly when $r>13$ or $r<-23$. We also want both of the solutions of the original quadratic equation to be negative. If $r>13$, then the equation $x^{2}-(r+7) x+r+87=0$ is of the form $x^{2}-b x+c=0$ with each of $b$ and $c$ positive. In this case, if $x<0$, then $x^{2}>0$ and $-b x>0$ and $c>0$ and so $x^{2}-b x+c>0$. This means that, if $r>13$, there cannot be negative solutions. Thus, it must be the case that $r<-23$. This does not guarantee negative solutions, but is a necessary condition. So we consider $x^{2}-(r+7) x+r+87=0$ along with the condition $r<-23$. This quadratic is of the form $x^{2}-b x+c=0$ with $b<0$. We do not yet know whether $c$ is positive, negative or zero. We know that this equation has two distinct real solutions. Suppose that the quadratic equation $x^{2}-b x+c=0$ has real solutions $s$ and $t$. This means that the factors of $x^{2}-b x+c$ are $x-s$ and $x-t$. In other words, $(x-s)(x-t)=x^{2}-b x+c$. Now, $(x-s)(x-t)=x^{2}-t x-s x+s t=x^{2}-(s+t) x+s t$. Since $(x-s)(x-t)=x^{2}-b x+c$, then $x^{2}-(s+t) x+s t=x^{2}-b x+c$ for all values of $x$, which means that $b=(s+t)$ and $c=s t$. Since $b<0$, then it cannot be the case that $s$ and $t$ are both positive, since $b=s+t$. If $c=0$, then it must be the case that $s=0$ or $t=0$. If $c<0$, then it must be the case that one of $s$ and $t$ is positive and the other is negative. If $c=s t$ is positive, then $s$ and $t$ are both positive or both negative, but since $b<0$, then $s$ and $t$ cannot both be positive, hence are both negative. Knowing that the equation $x^{2}-b x+c=0$ has two distinct real roots and that $b<0$, the condition that the two roots are negative is equivalent to the condition that $c>0$. Here, $c=r+87$ and so $c>0$ exactly when $r>-87$. Finally, this means that the equation $x^{2}-(r+7) x+r+87=0$ has two distinct real roots which are both negative exactly when $-87<r<-23$. This means that $p=-87$ and $q=-23$ and so $p^{2}+q^{2}=8098$.