Operations with surds (Pre-calculus) I have tried multiple ways to solve this:
$$ \left (2\sqrt 8 - 3\sqrt 5 \right)^2 $$
(original screenshot)
and my answer was way off. Can someone show me their method? Thanks.
 A: The simplest way to prevent any careless mistakes is to apply the identity
$$(a - b)^2 = a^2 + b^2 - 2ab$$
Substitute $a = 2\sqrt{8}$ and $b = 3\sqrt{5}$. Then we have 
$$a^2 = (2\sqrt8)^2 = 4\cdot8 = 32$$
and
$$b^2 = (3\sqrt{5})^2 = 9\cdot5 = 45$$
and
$$2ab = 2\cdot2\sqrt8\cdot3\sqrt5=12\sqrt{40}$$
So we have
$$(2\sqrt{8} - 3\sqrt{5})^2 = 32 + 45 - 12\sqrt{40} = 77 - 12\sqrt{40}$$
A: Remember the formula:
$$(a-b)^2=a^2-2ab+b^2$$
Apply it to your expression $(2\sqrt{8}-3\sqrt{5})^2$. But first, simplify $2\sqrt{8}$ to $4\sqrt{2}$. Now we have to find the value of:
$$(4\sqrt{2}-3\sqrt{5})^2$$
Use the $(a-b)^2=a^2-2ab+b^2$ formula.
$$a=4\sqrt{2}$$
$$b=3\sqrt{5}$$
$$(4\sqrt{2}-3\sqrt{5})^2=(4\sqrt{2})^2-2(4\sqrt{2})(3\sqrt{5})+(3\sqrt{5})^2$$
Remember, $(ab)^2=a^2b^2$.
$$(4\sqrt{2})^2-2(4\sqrt{2})(3\sqrt{5})+(3\sqrt{5})^2=4^2(\sqrt{2})^2-2(4\sqrt{2})(3\sqrt{5})+3^2(\sqrt{5})^2$$
$$4^2(\sqrt{2})^2-2(4\sqrt{2})(3\sqrt{5})+3^2(\sqrt{5})^2=16(2)-2(4\sqrt{2})(3\sqrt{5})+9(5)$$
$$16(2)-2(4\sqrt{2})(3\sqrt{5})+9(5)=32-2(4\sqrt{2})(3\sqrt{5})+45$$
$$32-2(4\sqrt{2})(3\sqrt{5})+45=77-2(4\sqrt{2})(3\sqrt{5})$$
Now we will simplify $2(4\sqrt{2})(3\sqrt{5})$.
$$77-2(4\sqrt{2})(3\sqrt{5})=77-2(4)(3)(\sqrt{2})(\sqrt{5})$$
Rememeber. $\sqrt{a}\cdot\sqrt{b}$.
$$77-2(4)(3)(\sqrt{2})(\sqrt{5})=77-24\sqrt{2\cdot 5}$$
$$77-24\sqrt{2\cdot 5}=77-24\sqrt{10}$$
$$\displaystyle \boxed{\therefore \left(2\sqrt{8}-3\sqrt{5}\right)^2=77-24\sqrt{10}}$$
