Geometry algebra

The expression should remind you of Heron's Formula, which is

$A = \sqrt{s(s-a)(s-b)(s-c)}$

where $a$ , $b$ , and $c$ are the sides of a triangle. and $s = \frac 1 2 (a+b+c)$ . Then, we can interpret this as finding the area of a triangle with sides $6$ , $10$ , and $\sqrt{136}$ . This happens to be a right triangle, so the area is $\frac{6 \times 10}{2} = \boxed{30}$ .

$\begin{aligned} \sqrt{x(10-x)(\sqrt{136}-x)(x-6)} & = \sqrt{(8+\sqrt{34})(2-\sqrt{34})(\sqrt{34}-8)(2+\sqrt{34})} \\ & = \sqrt{(\sqrt{34}+8)(\sqrt{34}-8)(2+\sqrt{34})(2-\sqrt{34})} & \small \blue{\text{Note that }(a+b)(a-b) = a^2 - b^2} \\ & = \sqrt{(34-64)(4-34)} \\ & = \sqrt{(-30)(-30)} \\ & = \boxed{30} \end{aligned}$