3x^2+4mx+m=0. I need a lot of help, the answer is x=-1/2 and m=3/4, but I never get that.

If m = 3/4 we have:
3x^2 + 4(3/4)x + (3/4) = 0
3x^2 + 3x + 0.75 = 0
x = (-3 +/- sqrt(3^2 - 4(3)(0.75))) / (2*3)
x = (-3 +/- sqrt(9 - 9)) / 6
x = (-3 +/- sqrt(0)) / 6
x = (-3 +/- 0) / 6
x = -3/6
x = -1/2

If x = -1/2, we have:
3(-1/2)^2 + 4m(-1/2) + m = 0
3(1/4) - 2m + m = 0
(3/4) - m = 0
m = 3/4

You can't solve for both; you need one to find the other. UNLESS you know that there's only one solution for x, in which case the discriminant must be zero:
(4m)^2 - 4(3)(m) = 0
16m^2 - 12m = 0
4m(4m - 3) = 0
4m = 0 or 4m - 3 = 0
m = 0/4 or 4m = 3
m = 0 or m = 3/4
If m = 0 then x = 0
If m = 3/4 then x = -1/2