so, we have 80 = x² -16x.
first off, we start by grouping the terms with "x".
80 = (x² - 16x)
80 = (x² - 16x + [?]²) <--- we have a missing fellow.
so, the idea being, we need "some value" to get our perfect square trinomial, hmmm what could that be?
well, the tell-tale guy is the middle term, we know that 2(√ guy on the left)(√ guy on the right) is the middle term.
we know the square root of the (√x²) is just "x", and the guy on the right is [?], but we also know that if we multiply 2 times both we get 16x, so
[tex]\bf 2(x)\left(\boxed{?}\right)=16x\implies \boxed{?}=\cfrac{16x}{2x}\implies \boxed{?}=8[/tex]
aha! there's our missing fellow, so we need to add 8².
however, bear in mind, all we're doing is borrowing from our very good friend Mr Zero, 0.
so if we add 8², we also have to subtract 8².
[tex]\bf 80=(x^2-16x+8^2-8^2)\implies 80=(x^2-16x+8^2)-64
\\\\\\
144=(x^2-16x+64)\implies 144=(x-8)^2\implies \pm\sqrt{144}=x-8
\\\\\\
\pm 12 = x-8\implies \pm 12 + 8 = x\implies x=
\begin{cases}
20\\
-4
\end{cases}[/tex]
