We have $x,y$ belong to integers satisfying the condition $$6x^2 + y^2 + 6x = 3xy + 6y + x^2y.$$ We need to find the solution for which $y$ is maximal.
By trial I get $x=4$ , $y=30$. Which matches the answer. However I am looking for subjective approach. I made a quadratic with respect to $y$ but things got messy.
$$y^2 - y(6+x+x^2) + 6x+6x^2$$ We need to have discriminate to be perfect Square
$$ D = (x^2 + x + 6)^2 -4(6x+6x^2)$$ However this is in 4th degree which I don't know how to handle
Still $$ p^2 = (x^2 + x + 6)^2 -4(6x+6x^2) $$ Using $$ (x^2+ x + 6 - p)(x^2 +x+ 6 + p) = 24( x + x^2 )$$