If 'x' is real and $\frac{x^2-x+c}{x^2+x+2c}$ can take all real values find the interval in which $c$ lies.
My approach:
$y=\frac{x^2-x+c}{x^2+x+2c}$
$\forall x, y$ is real only if ${x^2+x+2c}\neq0$
Hence the equation ${x^2+x+2c}$ has no root.
$D<0$
$b^2-4ac<0$
$1-8c<0$
$c>1/8$
But the answer is $(-6,0)$.
I can't figure out why.