7. Given that $a$ is a real number, for all real numbers $x$, the quadratic function $y=x^{2}-4 a x+2 a+30$ is always non-negative, the range of the root of the equation $\frac{x}{a+3}=|a-1|+1$ is $\qquad$
7. $\left[-\frac{9}{4}, 0\right) \cup(0,15]$.
$\Delta=16 a^{2}-4(2 a+30) \leqslant 0$, we get $-\frac{5}{2} \leqslant a \leqslant 3$. When $-\frac{5}{2} \leqslant a \leqslant 1$ and $a \neq-2$, $x=(a+$
2) $[-(a-1)+1]=4-a^{2} \in\left[-\frac{9}{4}, 0\right) \cup(0,4]$; when $1<a \leqslant 3$, $x=(a+1)^{2}-1 \in$
$(3,15]$. Therefore, $a \in\left[-\frac{9}{4}, 0\right) \cup(0,15]$.