Regarding quadratic congruences in the form, $\displaystyle y^2 \equiv d (\bmod p) $ where p i s a prime

1. How exactly do you solve for quadratic congruences in this form?

This is an example from online: $\displaystyle y^2 \equiv 13 (\bmod 35) $ , then $\displaystyle y \equiv 15 (\bmod 35) $ and $\displaystyle y \equiv 38 (\bmod 35) $. I do not understand the jump from the first step to the next.

2. In this form, $\displaystyle ax^2 + bx +c \equiv 0 ( \bmod p) $:

From one website, it said to:

i.Let $\displaystyle y = 2ax + b $ and $\displaystyle d = b^2-4ac $

ii. Then get $\displaystyle y^2 \equiv d (\bmod p) $ which is the I have above

ii. Then substitute y = 2ax + b back into the solutions of $\displaystyle y^2 \equiv d (\bmod p) $ and solve for x.

*How do you solve for a linear congruence in this form (for example):

$\displaystyle 10x - 3 \equiv 15 ( \bmod 53) $

Thank you for reading. Any help is greatly appreciated.