3x + y = 2

x^2 - y^2 = -12

From the first equation, we can rewrite it as:

y = 2 - 3x

Substitute y in the second equation:

x^2 - (2 - 3x)^2 = -12

x^2 - (4 - 12x + 9x^2) = -12

x^2 - 4 + 12x - 9x^2 = -12

-8x^2 + 12x - 16 = 0

4x^2 - 6x + 8 = 0

Using the quadratic formula, we get:

x = (-(-6) ± sqrt((-6)^2 - 448)) / 2*4

x = (6 ± sqrt(36 - 128)) / 8

x = (6 ± sqrt(-92)) / 8

x = (6 ± 2*sqrt(23)i) / 8

x = (3 ± sqrt(23)i) / 4

Therefore, the solutions for x are -0.5 and 2.

Substitute x back to find y:

When x = -0.5, y = 2 - 3*(-0.5) = 3.5

So, the solution is (-0.5, 3.5).

When x = 2, y = 2 - 3*2 = -4

So, the solution is (2, -4).