Getting rid of the fraction helps:
\(7\sqrt{\left(2a\right)^2+1}\:-\:4a^2-1=2\sqrt{1+4a^2}+6\)
Then, we subtract 6 from both sides.
Doing so, we have all of the contants on one side. Now, square the equation:
\(\left(-4a^2+7\sqrt{\left(2a\right)^2+1}-7\right)^2=\left(2\sqrt{1+4a^2}\right)^2 \rightarrow \\ 16a^4+252a^2-56a^2\sqrt{4a^2+1}-98\sqrt{4a^2+1}+98 =4+16a^2\)
Subtract 16a^4 and 252a^2 and 98 from both sides, and simplify:
\(-56a^2\sqrt{4a^2+1}-98\sqrt{4a^2+1}=-16a^4-236a^2-94\)
\(\mathrm{Subtract\:}-16a^4-236a^2-94\mathrm{\:from\:both\:sides}\)
You'll get the following:
\(-56a^2\sqrt{4a^2+1}-98\sqrt{4a^2+1}+16a^4+236a^2+94=0\)
Then factoring yields:
\(2\left(-28a^2\sqrt{4a^2+1}-49\sqrt{4a^2+1}+8a^4+118a^2+47\right)\)
Thus, this equation equals 0, and you can divide by 2. Now we subtract \(8a^4+118a^2\) and 47 from both sides.
Factoring again, \(-7\sqrt{4a^2+1}\left(4a^2+7\right)=-8a^4-118a^2-47\)
If you expand these, you will get a sextic equation (equation with degree six variables).
Plugging those into Wolfram yields \(a=\pm\sqrt 2, a=\pm\frac{\sqrt 3}{2}, \text{and } a=\pm \frac{\sqrt 2\sqrt{67+9\sqrt{57}}}{4}.\)
However, when you plug the last possibility in, it doesn't work. So, your answer is sqrt 2.
