I finally understand simple congruences. Now how to solve a quadratic congruence? Now that I have plain old congruences, $19x\equiv 4 \pmod {141}$ for example, I am trying to wrap my brain around quadratic ones. My textbook shows how to tackle the aforementioned congruences, but not quadratic.
$$15x^2 + 19x\equiv 5 \pmod {11}$$
The book hints to show that would be equivalent to 
$$15x^2 + 19x + 6\equiv 0 \pmod{11}$$
I have no idea how they got that. I've looked at previous answers, but I need a dumbed down version.

Here is how to solve it generally. Every $\equiv$ here refers to equivalence $\!\bmod {p}$.
If $a\not\equiv 0$ with $p$ odd prime, then $$ax^2+bx+c\equiv 0\Leftrightarrow 4a^2x^2+4abx+4ac\equiv 0\Leftrightarrow (2ax+b)^2\equiv b^2-4ac$$
So for the congruence to have solutions, necessarily $b^2-4ac\equiv z^2$ for some $z\in\mathbb Z_p$.
$$\Leftrightarrow 2ax+b\equiv \pm z\Leftrightarrow x\equiv \frac{-b\pm z}{2a}$$

If $a\not\equiv 0$ with $p$ odd prime and $b^2-4ac\equiv z^2$ for some $z\in\mathbb Z_p$, then $$ax^2+bx+c\equiv 0\Leftrightarrow x\equiv \frac{-b\pm z}{2a}$$
If $a\equiv 0$ or $p=2$, the congruence is trivial, and if $\left(\frac{b^2-4ac}{p}\right)=-1$ (Legendre symbol, meaning $b^2-4ac$ is not a square $\!\bmod p$), $a\not\equiv 0,$ $p$ odd prime, it has no solutions.

In your case, $a\not\equiv 0$ with $p$ odd prime ($p=11, a\equiv 15, b\equiv 19, c\equiv 6$) and $b^2-4ac\equiv 1^2$, so we can use the theorem above:
$$15x^2+19x+6\equiv 0\Leftrightarrow x\equiv \frac{-19\pm 1}{2\cdot 15}\equiv\frac{3\pm 1}{-3}\equiv \{-4\cdot 3^{-1}, -2\cdot 3^{-1}\}\equiv \{6,3\}$$
Here I used $3(3^{-1})\equiv 1\equiv 12\Leftrightarrow 3^{-1}\equiv 4$.
