One variable equation

It all comes down to a quadratic equation: $\frac{a+56+a^2}{2018} + \frac{56+a+56^2}{2018} = \frac{44051}{2018}$ Multiply both sides by $2018$ : $a^2+2a+3248=44051$ Subtracting $44051$ from both sides: $a^2+2a-40803=0$ The coefficients are as follows: $x=1,\space y=2,\space z=-40803$ . Now writing the quadratic equation: $a_{12}=\frac{-y\pm\sqrt{\Delta}}{2x}$ Whereas $\Delta$ is the discriminant, $\Delta=y^2-4xz$ . $\Delta=2^2+4\cdot 40803=4\cdot 40804$ Now, substituting with the numerical values: $a_{12}=\frac{-2\pm\sqrt{4\cdot 40804}}{2}$ A nice observation to make our life easier: $40804=202^2$ : $a_{12}=\frac{-2\pm 2\cdot 202}{2} = -1\pm 202$ So $a$ is either $201$ or $-203$ , and therefore $2018+a$ is either $2219$ or $1815$ . Since we are looking for the minimal value, the correct answer is 1815 .