When Algebra is on the mod

$\begin{aligned} x^2 - 4x + 56 & \equiv 14\pmod {17} \\ x^2 - 4x + 42 & \equiv 0 \pmod {17} \\x^2 - 4x + 42 +17x & \equiv 0\pmod {17} \\ x^2 +13x + 42 & \equiv 0 \pmod {17}\\(x+6)(x+7) &\equiv 0\pmod {17}\\ \implies x &\equiv-6\pmod {17}\equiv 11\pmod {17} \color{#3D99F6}\text{ or}\\ x &\equiv-7\pmod {17}\equiv 10\pmod {17}\\ \text{Since we need the minimum } &\text{postive value for x, we have ,}\\ x&=\color{#EC7300}\boxed{\color{#333333}10}\end{aligned}$

$\begin{aligned} x^2 - 4x + 56 & \equiv 14 \text{ (mod 17)} \\ (x-2)^2 + 52 & \equiv 14 \text{ (mod 17)} \\ (x-2)^2 + 1 & \equiv 14 \text{ (mod 17)} \\ (x-2)^2 & \equiv 13 \text{ (mod 17)} & \small \color{#3D99F6} \text{Note that }(x-2)^2 > 30 \implies x > 7 \\ 64 & \equiv 13 \text{ (mod 17)} \\ \implies x & = \boxed{10} \end{aligned}$