Positive Voltage on inductor
When you apply a voltage across an inductor, the formula that describes the current flow is this: -
$$V_{APPLIED} = L\cdot\dfrac{di}{dt}$$
So, with 1 volt applied to a 1 henry inductor, the result is a current rising at 1 amp per second. No negative signs were harmed in this formula!
When you induce a voltage (with a changing flux) you put a negative sign in front of the voltage formula to remind us that any current drawn from the terminals will be in a direction that tends to reduce the flux that initially induced the voltage.

Given the comments below and the misconceptions about back-emf I have decided to add a few more words about back-emf. What some folk levitate towards is the belief that the back-emf causes the inductor current to ramp up. This is naïve. Consider a pure inductor with 1 volt applied across its terminals. A back-emf of 1 volt is produced and this appears to suggest that current can never flow. However, appearances can be deceptive because we know that current does flow. So, back-emf is a red herring - it exists but it doesn't stop current flowing as per the relationship described earlier in my answer.
But, this seems to be at odds with the back-emf exactly equalling the applied emf. But, consider what is happening; the back-emf and the applied voltage are both constant values and, they are across a zero ohm impedance hence, the current cannot be defined other than saying: -
$$\boxed{\text{The current is: }\hspace{1cm}\dfrac{0}{0}}$$
Why did I say that the impedance is zero?
Answer: it's an inductor and, the back-emf is induced in series with that inductor hence, with the applied voltage and back-emf voltage being equal, there is 0 volts across the &quot;true&quot; inductance. Zero volts means zero spectral content across that inductor and, it therefore has to be represented by 0 ohms.
In other words, you cannot use the value of applied voltage and back-emf to calculate the current through the inductor. The current rising or falling delivers the back-emf and that's it.