If $x$ is real and $4y^2+4xy+x+6=0$, then the complete set of values of $x$ for which $y$ is real, is:
$\text{(A) } x\le-2 \text{ or } x\ge3 \quad \text{(B) }  x\le2 \text{ or } x\ge3 \quad \text{(C) }  x\le-3 \text{ or } x\ge2 \quad \\ \text{(D) } -3\le x\le2 \quad \text{(E) } -2\le x\le3$
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 $x \ge3$ or $\fbox{A}$