Solving irrational inequalities: $\sqrt{x^2-7x+10}+9\log_4{x/8}\geq 2x+\sqrt{14x-20-2x^2}-13$ 
Solving the following inequalities: $$\sqrt{x^2-7x+10}+9\log_4{x/8}\geq 2x+\sqrt{14x-20-2x^2}-13$$    

I have been solving some questions on inequalities lately and I have come across this question, but I can't figure out a way to solve this...it contains both log and square root. Here's how I approached the question: 
$\sqrt{x^2-7x+10}+9\log_4{\frac{x}{8}}\geq 2x+\sqrt{14x-20-2x^2}-13$    
$\sqrt{x^2-7x+10}+9\log_4{x}-3/2\geq 2x+\sqrt{(-2)(x^2-7x+10)}-13$   
$(1-2i)\sqrt{x^2-7x+10}+\geq 2x-9\log_4{x}+3/2-13$    
$(1-2i)\sqrt{(x-5)(x-2)}+\geq 2x-9\log_4{x}-23/2$    
I don't know how to proceed further. It would be great if I could get a hint.

If $x \in \mathbb R $, then we have that the expressions $\sqrt {x^2-7x-10} \text {  and  } \sqrt {20+14x-2x^2}$ should be positive, that is, $\geq 0$. But one expression is $\sqrt {2}i $ times the other implying that both expressions are zero, that is, $x=2$ and $x=5$ can satisfy this inequality.
But however $x=5$ does not satisfy the inequality giving us $\boxed {x=2} $ is the only solution. Hope it helps. 
