$$12x+5y^2=0$$
Express the equation in terms of x
Note: x and y are integers of whole numbers
$$\frac{12x}{12}=\frac{-5y^2}{12}$$
$$x=\frac{-5y^2}{12}$$
$$If\ \ y\ =\ 1$$
$$then\ x\ =\frac{-5\left(1\right)^2}{12}=\frac{-5}{12}=-0.42$$
$$x\ is\ not\ an\ integer\ so\ y\ne1$$
$$If\ \ y\ =\ 2$$
$$then\ x\ =\frac{-5\left(2\right)^2}{12}=\frac{-20}{12}=-1.67$$
$$x\ is\ not\ an\ integer\ so\ y\ne2$$
$$If\ \ y\ =\ 3$$
$$then\ x\ =\frac{-5\left(3\right)^2}{12}=\frac{-45}{12}=-3.75$$
$$x\ is\ not\ an\ integer\ so\ y\ne3$$
$$If\ \ y\ =\ 4$$
$$then\ x\ =\frac{-5\left(4\right)^2}{12}=\frac{-80}{12}=-6.62$$
$$x\ is\ not\ an\ integer\ so\ y\ne4$$
$$If\ \ y\ =\ 5$$
$$then\ x\ =\frac{-5\left(5\right)^2}{12}=\frac{-125}{12}=-10.42$$
$$x\ is\ not\ an\ integer\ so\ y\ne5$$
$$If\ \ y\ =\ 6$$
$$then\ x=\frac{-5\left(6\right)^2}{12}=\frac{-180}{12}=-15$$
$$x\ is\ an\ integer\ so\ y=6\ or\ multiple\ of\ 6$$
$$and\ x=-15\ or\ multiple\ of\ 15$$

Answer =B