Write the following sets in roster from:

The given equation is:

$x^{2}+x-12=0$

$\Rightarrow x^{2}+4 x-3 x-12=0$

$\Rightarrow x(x+4)-3(x+4)=0$

$\Rightarrow(x-3)(x+4)=0$

$\Rightarrow x-3=0$ or $x+4=0$

$\Rightarrow x=3$ or $x=-4$

therefore, the solution set of the given equation can be written in roaster form as {3, -4}

So, J = {3, -4}