How can I use this equation to solve for $t$? The problem given is, "Using the equation $R = R_{20} [1 + a_{20} (t-20)] $, find the temperature at which the resistance will increase to 1 Ohm from a level of 0.8 Ohm at 20C", where $a_{20} = 0.00393$ and $R_{20} = 0.8$ (Ohm).

I have the answer from the book (83.61C), but I've failed to rearrange the equation to make $t$ the subject and it seems to imply I should be using the equation as is. Am I missing something?
Note you have: $R=R_{20}\big(1+a_{20}(t-20)\big)$. First, use the distributive property on the inside, so you have $R=R_{20}(1+a_{20}t-20a_{20})$. Now use the distributive property again: $R=R_{20}+R_{20}a_{20}t-20R_{20}a_{20}$. Now you want to solve for $t$. So move everything that does not have a $t$ to the same side: $R-R_{20}+20R_{20}a_{20}=R_{20}a_{20}t$. Now to solve for $t$, divide by its coefficient:
$$
t=\frac{R-R_{20}+20 R_{20}a_{20}}{R_{20}a_{20}}
$$
Now all that is left to do is plug in $R=1, R_{20}=0.8,a_{20}=0.00393$ to get $t \approx 83.61323$, as you can see here.
$$R = R_{20}[1+ a_{20}(t - 20)]$$

$$R = R_{20}[1 + a_{20}t - 20a_{20}]$$

$$R = R_{20} + a_{20}R_{20}t - 20R_{20}a_{20}$$

$$a_{20}R_{20}t = -R_{20} +20R_{20}a_{20} +R$$

$$t = \frac{-R_{20} +20R_{20}a_{20} + R}{a_{20}R_{20}}$$