Large elements of my answer is drawn from my own notes that I use for teaching General Chemistry II.
What is −0.693?
Radioactive decay processes follow first-order kinetics. A first-order reaction is one where the rate depends only on the concentration of one of the reactants raised to the first power. So consider the following reaction:
$$\mathrm{A} \longrightarrow \mathrm{products}$$
The rate can be expressed as the rate of change in the reactant concentration with a change in time such that
$$\mathrm{rate} = -\dfrac{\Delta[\mathrm{A}]}{\Delta t}$$
This simply means that [A] is being consumed (i.e. converted to products) as the reaction proceeds with time. The corresponding rate law can be written as
$$\mathrm{rate} = k[\mathrm{A}]$$
where $k$ is a rate constant. We can set these two expressions equal to each other to give
$$-\dfrac{\Delta[\mathrm{A}]}{\Delta t} = k[\mathrm{A}]$$
At this point we must apply a bit of calculus. Let us write the previous equation in differential form to give
$$-\dfrac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]~~~\mathrm{or}
~~~\dfrac{d[\mathrm{A}]}{[\mathrm{A}]} = -kdt$$
Integrate over $t=0$ to $t = t$ to give
$$\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t}\dfrac{d[\mathrm{A}]}{[\mathrm{A}]} = -k\int_0^t dt$$
which results in
$$\ln[\mathrm{A}]_t - \ln[\mathrm{A}]_0 = -kt$$
or
$$ \ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=-kt$$
where $[\mathrm{A}]_0$ is an initial concentration and $[\mathrm{A}]_t$ is the concentration at some time, $t$.
Now that we have our first-order integrated rate law, we can now discuss half-life. Half-life (denoted as $t_{\frac{1}{2}}$) is simply the time required for some reactant concentration to drop to half its original value (where the original value is simply [A]$_0$. So
$$t = t_{\frac{1}{2}}~~~\mathrm{when}~~~[\mathrm{A}]_t = \dfrac{1}{2}[\mathrm{A}]_0$$
Given this relationship, we can rearrange our first-order integrated rate law to solve for $t$ such that
$$\ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0} = -kt~~~\longrightarrow~~~
t = \dfrac{1}{k}\ln\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_t} $$
Therefore, we can make some simple substitutions into our first-order integrated rate law at $t = t_{\frac{1}{2}}$ to give
$$t_{\frac{1}{2}} = \dfrac{1}{k}\ln\dfrac{[\mathrm{A}]_0}{\frac{1}{2}[\mathrm{A}]_0} $$
We can then simplify this equation such that
$$t_{\frac{1}{2}} = \dfrac{1}{k}\ln2\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_0} ~~~\longrightarrow~~~
t_{\frac{1}{2}} = \dfrac{1}{k}\ln (2)
~~~\longrightarrow~~~
t_{\frac{1}{2}} = \dfrac{0.693}{k}$$
So to answer your question, the 0.693 value comes from the $\ln(2)$ term. As ringo pointed out below (and I will include here to be thorough), you can now solve your first-order integrated rate law for $t$ and substitute in your half-life equation such that
$$\ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=-kt
~~~\longrightarrow~~~
\dfrac{\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}}{-k}=t
~~~\longrightarrow~~~
\dfrac{\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}}
{
-\frac{0.693}{t_{\frac{1}{2}}}
}
=t
$$
or more cleanly...
$$t = \dfrac{\left(t_{\frac{1}{2}}\right)
\left(\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}\right)}
{-0.693}
$$
How do you get $N_0$?
You could get $N_0$ by examining the activity of carbon-14 in a relatively recent object or plant. Carbon-14 in a living plant is not expected to have decayed at all (since the half-life is over 5000 years!). Therefore you could use its activity for $N_0$. Of course this procedure relies on a few assumptions but carbon-14 dating is an approximation to begin with.
