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

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

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

2. In this form, $ax^2 + bx +c \equiv 0 ( \bmod p)$:
From one website, it said to:
i.Let $y = 2ax + b$ and $d = b^2-4ac$
ii. Then get $y^2 \equiv d (\bmod p)$ which is the I have above
ii. Then substitute y = 2ax + b back into the solutions of $y^2 \equiv d (\bmod p)$ and solve for x.

*How do you solve for a linear congruence in this form (for example):
$10x - 3 \equiv 15 ( \bmod 53)$

Thank you for reading. Any help is greatly appreciated.

2. Firstly
$y^2 \equiv 13 (\bmod 35)$ , then $y \equiv 15 (\bmod 35)$ and $y \equiv 38 (\bmod 35)$
is not true. Check it.

3. Originally Posted by whipflip15
Firstly
$y^2 \equiv 13 (\bmod 35)$ , then $y \equiv 15 (\bmod 35)$ and $y \equiv 38 (\bmod 35)$
is not true. Check it.
Ah, yes you right. Dyslexia :S It should be

$y^2 \equiv 13 (\bmod 53)$, then $y \equiv 15 ( \bmod 53)$ and $y \equiv 38 (\bmod 53)$