The set of real values of x satisfying the inequality $|x^2+x-6 | < 6$, is

$(-4, 3)$ $(-3, 2)$ $(-4, -3)∪(2, 3)$ $(-4, -1) ∪(0,3)$

$(-4, -1) ∪(0,3)$

We have, $|x^2 + x −6] < 6$ $⇒-6<x^2+x-6<6$ $⇒-6 <x^2+x-6$ and $x^2 + x-6<6$ $⇒x^2 + x > 0$ and $x^2+x-12 <0$ $⇒x (x + 1) > 0$ and $(x + 4) (x-3) <0$ $⇒x∈(-∞,-1)∪(0, ∞)$ and $-4 <x<3$ $⇒x∈(-4, -1)∪(0,3)$