smc

We treat the equation as a quadratic equation in $y$ for which the discriminant $$D=16x^2-16(x+6)=16(x^2-x-6)=16(x-3)(x+2)$$ For $y$ to be real $D\ge 0$. This inequality is satisfied when $x\le -2$ or $x\ge 3$ or $\boxed{x\le -2\text{ or }x\ge 3}$