7. If $A$ is the sum of the absolute values of all roots of the equation
$$
x=\sqrt{19}+\frac{91}{\sqrt{19}+\cdots \sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}
$$

Find $A^{2}$.
7. The fraction to the right of the equation can be simplified to $\frac{a x+b}{c x+d}$ $(a, b, c, d$ are real numbers), then the equation is a quadratic equation, with at most two roots. H $x=\sqrt{19}+\frac{91}{x}$ has roots that are also $x=\frac{\sqrt{19}-\sqrt{38}}{2}$, which are the only two roots of the original equation. Therefore, $A=\sqrt{383}$, i.e., $A^{2}=383$.