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.

$$15x^2+19x\equiv 5 \mod 11$$ 
is equivalent to 
$$15x^2+19x +6 \equiv 0 \mod 11$$
 by the fact that we can subtract $5$ from both sides of the congruence (and $-5\equiv 6 \mod 11$).  
From here, we factor $15x^2+19x+6$ to give $(3x+2)(5x+3)$.  This means that our original quadratic congruence is equivalent to 
$$(3x+2)(5x+3)\equiv 0 \mod 11.$$
This will only hold true if 
$$3x+2 \equiv 0 \mod 11 \quad \text{or} \quad 5x+3 \equiv 0 \mod 11,$$
 or equivalently 
$$3x\equiv 9 \mod 11 \quad \text{or} \quad 5x\equiv 8 \mod 11.$$  
Since you're familiar with solving linear congruences, you should be able to solve these.
