Find the zeros of y=x^2-8x-3 by completing the square

Zeroes are where it equals 0

0=x^2-8x-3

group x terms

0 = (x^2-8x) - 3

take 1/2 of linear coefient and square it

-8/2=-4, (-4) ^2=16

add that to both sides

16 = (x^2-8x+16) - 3

factor

16 = (x-4) ^2-3

add 3 to both sides

19 = (x-4) ^2

sqrt both sides

+/-√19=x-4

add 4

4+/-√19=x

the zeroes are at x=4+√19 and 4-√19