2x^2-7x-2=0, Is This Right 7+/-sq Rt 53/4?

Let's see if your answer works.

2x² - 7x - 2 = 0
2(7+√(53/4))² - 7(7+√(53/4)) - 2 =? 0    (substitute 7+√(53/4) for x)
2(7² + 2*7√(53/4) + 53/4) - 49 - 7√(53/4) - 2 =? 0    (offhand, I'd say it doesn't look good)
98 + 28√(53/4) + 13 1/4 - 49 - 7√(53/4) - 2 =? 0
60 1/4 + 21√(53/4) ≠ 0    (your proposed answer is incorrect)
_____
ax²+bx+c = 0 has answer
x = (-b ±√(b²-4ac))/(2a)
x = (-(-7)±√((-7)2-4(2)(-2)))/(2*2)
  = (7 ±√(49+16))/4
x = (7±√65)/4

Please note the parentheses--missing from your problem statement. Note that the missing parentheses resulted in your question being misinterpreted. Apparently, you meant (7±√53)/4, rather than 7±√(53/4) or 7±(√53)/4

Let's try this answer.
2((7+√65)/4)² -7((7+√65)/4) - 2 = 0
2(49 + 14√65 + 65)/16 + (-49 -7√65)/4 - 2 = 0
(114 + 14√65)/8 - 49/4 - (7/4)√65 - 2 = 0
14 1/4 - 12 1/4 - 2 + (√65)(14/8 - 7/4) = 0
0 = 0    (yes)