Please help...

x^2/4 − y^2/2 = 1

x^2 = 4(y+2)        (1)

Multiply the first equation through by  4  and we get that

x^2 -  2y^2  =  4

x^2  = 2y^2 + 4      (2)

Equate (1) and (2)   and we have that

4(y + 2)   =  2y^2 + 4

2 (y + 2)  = y^2 + 2

2y + 4  =  y^2 + 2

y^2 - 2y - 2  =  0      complete the square on y

y^2 - 2y + 1   - 2  - 1   = 0

( y - 1)^2  - 3  =  0

(y - 1)^2  =  3

y - 1  = ±√3

y = 1 ±√3

y = 1 + √3      or  y  = 1 -√3

And

x^2  = 4 ( y + 2)

x = ±2√[ y + 2]

So the solutions are

(2√[ 3 - √3] , 1 -√3 )  

(-2√[ 3 - √3] , 1 -√3 )

(2√[ 3 + √3] , 1 +√3 )

(-2√[ 3 + √3] , 1 +√3 )