x^2 -  2x - 4y^2  +  8y -  2 = 0    (1)

5x^2 - 10x + 12y^2 + 24y - 58 = 0    (2)

Look at the system. You see that the two terms with  x^2 and x  of the second equation are five times 
the corresponding terms of the first equation.


It gives me an idea to multiply first equation by 5 and then subtract from the second equation
to ELIMINATE all the terms with x.


So my first step is to multiply first equation by 5, while keeping second equation as is


    5x^2 - 10x - 20y^2 + 40y - 10 = 0     (1')

    5x^2 - 10x + 12y^2 + 24y - 58 = 0     (2)


My second step is to subtract equation (1') from equation (2).  In this way, I get


                 32y^2 - 16y - 48 = 0.


Next, use the quadratic formula


     =  =  = .


So, I have two solutions for y:   = -1  and   =  = .


The last step is to substitute these found values of y into equation (1) and solve it for x.


It is just fully mechanical job and I leave it to you.


You should find two values of x for each of the two values of y.


It will give you four final solutions for pairs (x,y) at the end.