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