To solve the first equation 7x^2 + 2x - 3 = 0, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 7, b = 2, and c = -3. Plugging in these values, we get:
x = (-2 ± √(2^2 - 47(-3))) / 2*7
x = (-2 ± √(4 + 84)) / 14
x = (-2 ± √88) / 14
x = (-2 ± 2√22) / 14

Therefore, the solutions to the first equation are:
x = (-2 + 2√22) / 14
x = (-2 - 2√22) / 14

To solve the second equation 4x^2 - 6x + 12 = 0, we can also use the quadratic formula:
In this case, a = 4, b = -6, and c = 12. Plugging in these values, we get:
x = (-(-6) ± √((-6)^2 - 4412)) / 2*4
x = (6 ± √(36 - 192)) / 8
x = (6 ± √(-156)) / 8

The square root of a negative number means that the equation has complex solutions, so the solutions to the second equation are:
x = (6 ± √156i) / 8