Solve the following system of equations:

                         3x² + 4xy - 2y² = 2 
                                 3x + 2y = -10 

Solve the second equation for either of the unknowns.
I'll pick x:

                                 3x + 2y = -10

add -2y to both sides

                                      3x = -10 - 2y

Divide both sides by 3
             
                                       x = (-10 - 2y)/3                                  
factor out -2

                                       x = -2(5 + y)/3
           
In the first original equation,

                         3x² + 4xy - 2y² = 2

substitute -2(5 + y)/3 for x

 3[-2(5 + y)/3]² + 4[-2(5 + y)/3]y - 2y² = 2

  3[4(5 + y)²/9] + 4y[-2(5 + y)/3] - 2y² = 2

         4(5 + y)²/3 - 8y(5 + y)/3 - 2y² = 2

4(25 + 10y + y²)/3 - (40y + 8y²)/3 - 2y² = 2

Multiply through by 3 to clear of fractions:

    4(25 + 10y + y²) - (40y + 8y²) - 6y² = 6

       100 + 40y + 4y² - 40y - 8y² - 6y² = 6

Collect terms 

                              100 - 10y² = 6

Get 0 on the right

                               94 - 10y² = 0

Divide thru by 2                47 - 5y² = 0  

                                    -5y² = -47

                                      y² = 47/5
                                             ____
                                       y = ±Ö47/5
                                             ___
Rationalize denominator                y = ±Ö235/5  


Substitute into                  3x + 2y = -10 


Using the + value
                                 ___
                         3x + 2(Ö235/5) = -10

Multiply through by 5
                                     ___
                             15x + 2Ö235 = -50  
                                           ___
                             15x = -50 - 2Ö235                                
                                            ___
                               x = (-50 - 2Ö235)/15
                                            ___ 
                               x = -2(25 + Ö235)/15                

So one solution is 
                    ___       ___
(x, y) = ( -2(25 + Ö235)/15, Ö235/5 )

----------------------------------------             

Using the - value
                                  ___
                         3x + 2(-Ö235/5) = -10

Multiply through by 5
                                     ___
                             15x - 2Ö235 = -50  
                                           ___
                             15x = -50 + 2Ö235                                
                                            ___
                               x = (-50 + 2Ö235)/15
                                            ___ 
                               x = -2(25 - Ö235)/15                

So the other solution is 
                    ___        ___
(x, y) = ( -2(25 - Ö235)/15, -Ö235/5 )

Tell your teacher I said he or she is a meany for giving you such a 
messy problem.  It is messy enough as it is. However, if you solve for
y first instead of x, the problem is even messier, in fact a lot
messier.  That was the first way I did it!!  Also there is no way to
tell in advance which way is going to be messier.  Your teacher needs
a spanking!!  :-)

Edwin