Can you solve this problem?

$\large\begin{aligned}5^{1+\sqrt{x^2-4x-1}}+5^{\frac{5+4x-x^2}{2+\sqrt{x^2-4x-1}}}=&\ 126\\5^1\cdot5^{\sqrt{x^2-4x-1}}+5^{\frac{5+4x-x^2}{2+\sqrt{x^2-4x-1}}\cdot\frac{2-\sqrt{x^2-4x-1}}{2-\sqrt{x^2-4x-1}}}=&\ 126\\5\cdot5^{\sqrt{x^2-4x-1}}+5^{\frac{(5+4x-x^2)(2-\sqrt{x^2-4x-1})}{4-(x^2-4x-1)}}=&\ 126\\5\cdot5^{\sqrt{x^2-4x-1}}+5^{\frac{(5+4x-x^2)(2-\sqrt{x^2-4x-1})}{5+4x-x^2}}=&\ 126\\5\cdot5^{\sqrt{x^2-4x-1}}+5^{2-\sqrt{x^2-4x-1}}=&\ 126\\5\cdot5^{\sqrt{x^2-4x-1}}+\frac{5^2}{5^{\sqrt{x^2-4x-1}}}=&\ 126\end{aligned}$

Let $\large a=5^{\sqrt{x^2-4x-1}}$ That is,

$\begin{aligned}5a+\frac{25}a=&\ 126\\5a^2+25=&\ 126a\\5a^2-126a+25=&\ 0\\(5a-1)(a-25)=&\ 0\\a=&\ \frac15,\ 25\end{aligned}$

Case $(1)$ : $a=\frac15$ That is,

$\large\begin{aligned}5^{\sqrt{x^2-4x-1}}=&\ \frac15\\5^{\sqrt{x^2-4x-1}}=&\ 5^{-1}\\\sqrt{x^2-4x-1}=&\ -1\end{aligned}$

But $\sqrt{x^2-4x-1}\geq0$ Thus, this case has no real solution.

Case $(2)$ : $a=25$ That is,

$\large\begin{aligned}5^{\sqrt{x^2-4x-1}}=&\ 25\\5^{\sqrt{x^2-4x-1}}=&\ 5^2\\\sqrt{x^2-4x-1}=&\ 2\\x^2-4x-1=&\ 4\\x^2-4x-5=&\ 0\\(x+1)(x-5)=&\ 0\\x=&\ -1,5\end{aligned}$

Hence, $A=\{-1,5\}$ and the sum of the elements in set $A$ is $-1+5=\boxed4$