e^(2x)-6e^x+5=0 has two solutions, what are they?

I tried this:

1. moved +5 to the other side so there wouldnt be a ln 0 for the next step

2. applied ln to both sides [2x-6x=ln(-5)]

3. combined like terms = -4x=ln(-5)

4. isolated x = x=ln(-5)/-4

obviously this is wrong

the only thing i can think of that would yield two answers is the quadratic formula, but i don't know if thats the right route or how to set that up

edit still stuck here. any tips on where to start?

edit2: Ok. I just looked at this problem and realized that e^0 would work for one answer. I assume the other I can work out now.

edit3: still stuck. tried everything i know