To solve this system of equations, we can use the method of substitution or elimination.

Solve the first equation for y:
x^2 - y = 14
y = x^2 - 14

Substitute the expression for y into the second equation:
3x + (x^2 - 14) = 4
x^2 + 3x - 10 = 0

Factor the quadratic equation:
(x + 5)(x - 2) = 0

Set each factor equal to zero and solve for x:
x + 5 = 0 or x - 2 = 0
x = -5 or x = 2

Substitute the values of x back into the first equation to solve for y:
For x = -5:
y = (-5)^2 - 14
y = 25 - 14
y = 11
So one solution is x = -5, y = 11

For x = 2:
y = 2^2 - 14
y = 4 - 14
y = -10
So another solution is x = 2, y = -10

Therefore, the solutions to the system of equations are x = -5, y = 11 and x = 2, y = -10.