This is by far one of the most exaggerated problems I've seen. I saw it in a ReactionTime video(of all places) and I decided to attempt it. I have a few solutions where stuff gets equated to 0 but I'm not sure how to proceed from there.
Here it is: Solve for x in
$\displaystyle\dfrac{6(3x^2-27) \cdot 8x^2}{4(9-3x)(x^2+3x)}=\dfrac{\tan(x+4)}{\log_{10}(x+\frac{1}{4})}$
It's not looking good but at least some intense factorization is possible.
$\displaystyle\dfrac{6\cdot3(x+3)(x-3) \cdot 8x^2}{-4\cdot3(x-3)\cdot x(x+3)}=\dfrac{\tan(x+4)}{\log_{10}(x+\frac{1}{4})}$
$\displaystyle 144x^2(x+3)(x-3) \cdot \log_{10}(x+\frac{1}{4})=-12x(x+3)(x-3) \cdot \tan(x+4)$
$\displaystyle 144x^2(x+3)(x-3) \cdot \log_{10}(x+\frac{1}{4})+12x(x+3)(x-3) \cdot \tan(x+4)=0$
$\displaystyle 12x(x+3)(x-3)(12x \cdot \log_{10}(x+\frac{1}{4})+\tan(x+4))=0$
So my actual question was: how do you solve
$12x \cdot \log_{10}(x+\frac{1}{4})+\tan(x+4)=0$
Please help!