Time required store half the maximum energy in an inductor

[prodo123]

Homework Statement

A 35.0 V battery with negligible internal resistance, a 50 Ω resistor and a 1.25 mH inductor forms a RL circuit. How long will it take for the energy stored in the inductor to reach one-half of its maximum value?

Homework Equations

$i(t)=I_0 e^{\frac{-tR}{L}}$
$U(t)=\frac{1}{2}Li(t)^2$

The Attempt at a Solution

$U=\frac{1}{2}Li^2=\frac{1}{2}U_0$
$\frac{1}{2}Li^2=\frac{1}{4}LI_0^2$
$i^2=\frac{1}{2}I_0^2$
$i=I_0 e^{\frac{-tR}{L}}=\frac{1}{\sqrt{2}}I_0$
$\frac{-tR}{L}=\text{ln}(\frac{1}{\sqrt{2}})=-\frac{1}{2}\text{ln}(2)$
$t=\frac{L}{2R}\text{ln}(2)$
$t=8.66\text{ µs}$

Textbook has the following answer:
$t=30.7\text{ µs}$

Am I doing something wrong? The same method worked for finding how long it takes for the current to reach half the maximum value $I_0=\frac{\mathcal{E}}{R}$.

 

[prodo123]

Found solution slides here: http://clas.sa.ucsb.edu/docs/defaul...physics-4-inductanceE0239A696CF5.pdf?sfvrsn=2

Used the wrong current formula. Correct formula is:
$i(t)=I_0(1- e^{\frac{-tR}{L}})$