Find the interval in which 'm' lies so that the expression

$\displaystyle \frac{mx^2+3x-4}{-4x^2+3x+m}$

can take all real values ,where x is real.

i have equated this equation to y

$\displaystyle y=\frac{mx^2+3x-4}{-4x^2+3x+m}$

then

$\displaystyle (-4y-m)x^2+(3y-3)x+(ym-4)=0$

but then it becomes quadratic equation in many variables .

please give hint or steps to solve the problem