Welcome 2020

We need to find

$\begin{aligned} x^2 + 20 x - 380 & \equiv 0 \text{ (mod 2020)} \\ (x+10)^2 - 480 & \equiv 0 \text{ (mod 2020)} \\ (x+10)^2 & \equiv 480 \text{ (mod 2020)} \\ \implies 2020 \blue n + 480 & \equiv (x+10)^2 & \small \blue{\text{where }n \text{ is an integer}} \end{aligned}$

By inspection: When $n=1 \implies 2020+480 = 2500 = 50^2 = (x+10)^2 \implies x = \boxed{40}$ . Note that $n=1$ is the smallest $n$ , hence the smallest LHS, which the solution exists, $x = 40$ is the smallest $x$ .

By reasoning: Since the LHS, $2020n + 480 = 20(101n+24)$ is divisible by 20, and RHS is a square, the RHS must be divisible by $2^25^2 = 100$ . This also means that the LHS is also divisible by 100. Then we have:

$\begin{aligned} 2020n + 480 & \equiv 0 \text{ (mod 100)} \\ 20n + 80 & \equiv 0 \text{ (mod 100)} \\ \implies n & \equiv 1 \\ \implies x & \equiv \boxed{40} & \small \blue{\text{Again the smallest solution.}} \end{aligned}$

We can write the given expression as $(x+10)^2-480$ . So the minimum of this expression is $2020$ . Therefore $(x+10)^2=2020+480=2500$ or $x=\sqrt {2500}-10=50-10=\boxed {40}$ (since $x>0$ )

$x^2 +20 x - 380 = 2020$

$x(x+20) = 2400 = 40 \times 60$

$x = 40$