Can someone help me solve by completing the square? 2x^2-7x-4=0

I have solved it by using the quadratic formula as well as by factoring and the zero product property but can't seem to figure out completing the square. Thank you!

2x^2-7x-4=0
2(x² - 7/2x) = 4
(x² - 7/2x) = 2
(x² - 7/2x + 49/16) = 2 + 49/16 = 81/16
(x-7/4)² = (9/4)²
x = 7/4 ± 9/4
x = -1/2, 4

(2x+1) (x-4) = 0
2x+1 = 0 (OR) x-4=0
x= - 1/2 (OR) x= 4

y=(x-3.5)^2 -11