Answer
The term containing $x^2$ in the expansion is $75x^2$.
Work Step by Step
Using Binomial Theorem,
$(\sqrt{x}-\sqrt{5})^6$ =
$(\sqrt{x})^6$ +
$\frac{6}{1!}(\sqrt{x})^{6-1}(-\sqrt{5})^1$ +
$\frac{6(6 -1)}{2!}(\sqrt{x})^{6-2}(-\sqrt{5})^2$ +
$\frac{6(6-1)(6-2)}{3!}(\sqrt{x})^{6-3}(-\sqrt{5})^3$ +
$\frac{6(6-1)(6-2)(6-3)}{4!}(\sqrt{x})^{6-4}(-\sqrt{5})^4$ +
$\frac{6(6-1)(6-2)(6-3)(6-4)}{5!}(\sqrt{x})^{6-5}(-\sqrt{5})^5$ +
$(-\sqrt{5})^6$
And, the term containing $x^2$ in the expansion is
= $\frac{6(6 -1)}{2!}(\sqrt{x})^{6-2}(-\sqrt{5})^2$
= $\frac{(6 \times 5)}{(2 \times 1)}(\sqrt{x})^4(5)$
= $75x^2$
