I am confused on how to solve this question:

**Prove that 3x**

^{2}-2mx+m-1 = 0 has real solutions for all real values of mFrom my understanding, "Real solutions" suggests that I should use the discriminant, for when b

^{2}- 4ac is either = or > 0.

So, this is where a = 3, b = -2m and c = m-1,

and so: (-2m)

^{2}- 4(3)(m-1) > 0, or (-2m)

^{2}- 4(3)(m-1)

**=**0

also, -4m

^{2}-12m + 12 > 0

From here, I am stuck, as solving this quadratic will not answer the original question.

Is this actually the way to prove this, or have I made a mistake.

Thanks