Why do we assume that this quadratic has one solution?

When you insert $y=mx+12$ into the circle equation and obtain $$(1+m^2)\cdot x^2 + (14m−12)\cdot x + 68 = 0\label{1}\tag{1}$$ There are two things you can do: solve for $x$ or for $m$. Which of those makes sense?

Notice that in $(\ref{1})$, if you take $x$ as a given constant and solve for $m$, you are answering the question: Given an $x$ intercept of the line through $(0,12)$ with the circle, determine the slope of this line. It might happen that there are two points, one or none on the circle with that given $x$ coordinate, hence the number of solutions for $m$. This isn't, however, what we are looking for.

Instead, the tangent-condition offers us valuable information: there is only one $x$ solution for some $m$ we are looking for. Thus, it makes more sense to solve for $x$ in $(\ref{1})$ with the quadratic equation formula: $$x = \frac{- 7 m + 6\pm\sqrt{-19 m^2 - 84 m - 32} }{m^2 + 1}$$ This might look frightening at a first glance, but remember that we already know that there is exactly one solution for $x$, and this happens precisely when the discriminant is zero: $$\Delta=-19m^2-84m-32=0$$ Which can be solved via the quadratic equation formula. This yields $$\fbox{$\displaystyle m\in\left\{-4, -\frac{8}{19}\right\}$}$$

Just to restate the essential point so that it doesn't get buried:

• The solutions to the equation correspond to intersection points between the line and the circle. There can be zero, one or two such points.
• If the line does not touch the circle, there are no (real) solutions.
• If the line passes through the interior of the circle, there are two solutions.
• Only if the line is tangent to the circle, i.e. only touches it at one point, will the equation have only one solution. Since we want to find lines that are tangent to the circle, this is exactly the case that we're interested in.

Conveniently, we also (should) know that a quadratic equation has only one solution if and only if its discriminant is zero. So we don't actually need to solve the original equation — it's enough to find the value(s) of $m$ for which the discriminant is zero. But that's just an algebraic shortcut. If we didn't know this fact already, we could've obtained the same result slightly more laboriously by first applying the quadratic formula to find the (up to) two solutions, and then finding the value(s) of $m$ for which they are equal (i.e. actually just one solution).