Basic Algebra Math Problem. Can't figure it out. Help please?!?!?!?

x^2 = 9y + 54, y = x-2 or x =y + 2
(y+2)^2 = 9y + 54
y^2 + 4y +4 = 9y + 54
y^2 - 5y - 50 = 0
(y-10)(y+5) = 0
y - 10 = 0, y+5 = 0
y = 10,-5
When y = 10, x = 10 + 2 =12
When y = -5, x =-5 + 2 =3
Solutions are x=12, y=10 (12,10) and x=-3,y=-5 (-3,-5)

x^2 = 9(x - 2) + 54
x^2 = 9x - 18 + 54
x^2 - 9x - 36 = 0
(x + 3)(x - 12) = 0
x = -3, 12
y = -5, 10

(-3, -5) (12, 10)

x² = 9 [ x - 2 ] + 54
x² = 9x + 36
x² - 9x - 36 = 0
[ x - 12 ] [ x + 3 ] = 0
x = 12 , x = - 3
y = 10 , y = -5

Substitute given y into second equation and solve for x then plug in to find y

x^2 = 9(x - 2) + 54

x^2 = 9x - 18 + 54

x^2 = 9x + 36

x^2 - 9x - 36 = 0

(x - 12)(x + 3) = 0

x = 12 and x = -3

y = x - 2
y = -3 - 2
y = -5

y = x - 2
y = 12 - 2
y = 10

So you have two solutions here..x = -3, y = -5 and x = 12, y = 10