x²-8x=3
we start off by grouping the "x"s, so
(x² - 8x) = 3
(x² - 8x + [?]²) = 3
now, we have a missing fellow, and the idea being, we use the middle term to get it, since we know that in a perfect square trinomial the middle term is just a product of left and right guys times 2, then,
![\bf 2(x)[?]=8x\implies [?]=\cfrac{8x}{2x}\implies [?]=4](https://img.qammunity.org/2019/formulas/mathematics/middle-school/wb6n6dq72qe8fmcdpiot2nkuk6pdqxe591.png)
so our missing fellow is 4, however all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add 4², we also subtract 4².
(x² - 8x +4² - 4²) = 3
(x² - 8x + 4²) - 16=3
(x - 4)² = 19
x - 4 = ±√(19)
x = 4±√(19)