Find the relationship of the length of triangle's sides. 
Denote the three sides of $\triangle ABC$ to be $a,b,c$. And they satisfy $$a^2+b+|\sqrt{c-1}-2|=10a+2\sqrt{b-4}-22 $$
  Now determine what kind of triangle $\triangle ABC$ is.
  A.Isosceles triangle which its leg and base is not equal.
  B.equilateral triangle
  C.Right triangle
  D.Isosceles Right triangle

The only information I got is from the number in the radical need to be greater than $0$. Then $b\ge4$ and $c\ge 1$. Also $10a+2\sqrt{b-4}-22\ge0 $. But they are all inequalities. What we need is some equalities. It would be great to have some hints.

HINT:
Write the equality as $$a^2-10a+22+b+|\sqrt{c-1}-2|-2\sqrt{b-4}=0$$ and since we know that $a$ is real, $$22+b+|\sqrt{c-1}-2|-2\sqrt{b-4}\le25\\b-2\sqrt{b-4}\le3-|\sqrt{c-1}-2|\le3.$$ But if $f(b)=b-2\sqrt{b-4}$, $f'(b)=1-\dfrac1{\sqrt{b-4}}=0$ for stationary points, resulting in $b=5$ as a minimum, and $f(b)=3$.
Hence that is the only value for $b$, meaning that $c=\cdots\,\,?$

Spoiler:

 The triangle is equilateral.

