Solve the quadratic congruences, or show them to be unsolvable. 
Solve the following congruences, or show them to be unsolvable:

$(a) 3x^2 - 5x +7 \equiv 0 \mod 13$
Since $9 \cdot 3 \equiv 1 \mod 13$, $9 \cdot 5 \equiv 6 \mod 13$ and $9 \cdot 7 \equiv 11 \mod 13$ we can multiply the above congruence by $9$ to obtain $x^2 - 6x + 11 \equiv 0 \mod 13$. Completing the square, we obtain $x^2 - 6x + 9 = (x-3)^2 \equiv -2 \mod 13$. Since the congruence $x^2 \equiv -2 \mod 13$ has no solutions, it follows that $(x-3)^2 \equiv -2 \mod 13$ has no solutions. $\textbf{Is this correct?}$
$(b) 5x^2 - 6x + 2 \equiv 0 \mod 13$
Since $8 \cdot 5 \equiv 1 \mod 13, 8 \cdot 6 \equiv 9 \mod 13,$ and $16 \equiv 3 \mod 13$, we can multiply the congruence $(b)$ by $8$ to obtain $x^2 - 9x +3 \equiv 0 \mod 13$. Completing the square, we obtain $(x - \frac{9}{2})^2 \equiv \frac{69}{4} \mod 13$. $\textbf{How do I proceed from here?}$

About b). Yes multiplication by 8 is correct. The polynomial you get is the same as $ x^2 +4 x + 3 \equiv 0 \pmod {13} $. Clearly you get $(x+1)(x+3) \equiv 0 \pmod {13} $. So $ x=10$ or $ x=12 \pmod {13} $.
