The given equation is:

${x}^{2}+x\u201312=0$⇒

${x}^{2}$ + 4x – 3x – 12 = 0

⇒ x(x + 4) – 3(x + 4) = 0

⇒ (x – 3)(x + 4) = 0

⇒ x – 3 = 0 or x + 4 = 0

⇒ 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}