How do I solve this quadratic simultaneous equation 2y+2x=7 and x^2-4y^2=8 it has two sets of solutions and I'm solving by both variables

These are all the possible answers i know from experience.

2x + 2y = 7

x^2 - 4y^2 = 8

2x = - 2y + 7

x = - y + 7/2

(-y + 7/2) ^2 - 4y^2 = 8

y^2 - 7y + 49/4 - 4y^2 = 8

0 = 3y^2 + 7y + 8 - 49/4

0 = 12y^2 + 28y + 32 - 49

0 = 12y^2 + 28y - 17

y = (-b ± √ (b^2 - 4ac)) / (2a)

y = (-28 ± √ (28^2 - 4 (12) (-17))) / (2 (12))

Dividing by 2 is like dividing by √ (4):

y = (-14 ± √ (7*28 + (12) (17))) / (2*6)

y = (-7 ± √ (7*7 + (3) (17))) / 6

y = (-7 ± 10) / 6 = 1/2 or - 17/6

x = - y + 7/2 = 3 or 19/3

Solutions are (3,1/2) and (19/3,-17/6)