Find L in $T=2\pi\sqrt{\frac{L}{g}}$ I am given the formula for the period of a pendulum: $T=2\pi\sqrt{\frac{L}{g}}$ along with inputs $T=1$ and $g=32$.

I am to find L, the textbook provides the answer as 10 (to the nearest whole unit). I arrive at 0.8.

I'm not sure I approached this the right way initially. How can I arrive at 10?

$1=2\pi\sqrt{\frac{L}{4\sqrt{2}}}$

$1^2=4\pi^2\frac{L}{32}$  # square both sides to get rid of radical

$\frac{1}{1}=\frac{4\pi^2L}{32}$  # writing the left side a a fraction just makes me see clearer when working with fraction equations

$32=4\pi^2 L$  # multiple out denominator

$32=39.5L$  # $4*(3.14159^22)$ ~ 39.5

$L = 32/39.5=0.8$

How can I arrive at 10?

Here's a screen shot from the textbook in case I've missed anything. It's the second question starting "If the gravity is 32..."
You got an answer of $0.8$. But since the question gave gravity in feet, your final answer is in feet! Since 1 foot is 12 inches, your final answer would be $$0.8\times 12=9.6\,\text{inches}$$
$L= g ({ T \over 2 \pi})^2 = 32{1 \over 4 \pi^2}= { 8 \over \pi^2} \approx {8 \over 9} \approx 1$.