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://tex.z-dn.net/?f=%5Cbf%202%28x%29%5B%3F%5D%3D8x%5Cimplies%20%5B%3F%5D%3D%5Ccfrac%7B8x%7D%7B2x%7D%5Cimplies%20%5B%3F%5D%3D4)
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)