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.

$0\equiv 4x^2\!+\!8x\!-\!5\equiv \overbrace{(2x)^2\!+\!4(2x)\!-\!5}^{\large X^2\ +\,\ 4\,X\,\ -\,\ 5\!\!\! }\equiv\overbrace{(2x\!+\!5)(2x\!-\!1)}^{\large (X\ +\ 5)\ (X\ -\ 1)}\,$ so $\,\begin{align} &2x\equiv\color{#c00}{-5}\equiv6\\ &2x\equiv\,\color{#c00} 1\,\equiv 12\end{align}\ $ so $\ x\equiv \ldots$

Or, complete the square $\,0\equiv 4x^2\!+\!8x\!-\!5\equiv (2x\!+\!2)^2\!-\!3^2$ so $\,2x\!+\!2\equiv \pm3\,$ so $\,2x\equiv\color{#c00}{ -5,1}\dots$

Or, apply the quadratic formula. First we make the lead coef $=1$ by scaling by $1/4\equiv 12/4\equiv3\,$ to get $\ 3(4x^2 + 8x - 5)\equiv x^2 + \color{#0a0}2\,x-4\equiv 0.\,$ It has discriminant $\, 2^2\!-4(-4)\equiv 20\equiv 9\equiv  \color{#a0f}3^2,\,$ therefore the roots are $\,x\equiv(-\color{#0a0}2\pm\color{#a0f} 3)/2\equiv \{\color{#c00}{-5,1}\}/2\equiv \{6,12\}/2\equiv \{3,6\}$
