Find the interval in which $m$ lies so that the expression $\frac{mx^2+3x-4}{-4x^2+3x+m}$ can take all real values, $x$ being real 
Find the interval in which $m$ lies so that the expression $\frac{mx^2+3x-4}{-4x^2+3x+m}$ can take all real values, $x$ being real.    

I don't know how to proceed with this question. I have equated this equation with $y$ to obtain a quadratic equation: $(m+4y)x^2+(3-3y)x-(4+my)=0$. Now I have no idea as to how I can find the answer. A small hint will be helpful.

Hint
$$f(x)=\frac{mx^2+3x-4}{-4x^2+3x+m}$$
The function $f$ must to be onto. It means that:
$$p=\frac{mx^2+3x-4}{-4x^2+3x+m} \Rightarrow (m+4p)x^2+3x(1-p)-4-pm=0$$
The above equation must have roots for any $p \in \Bbb R$. It means that:
$$\Delta=9(1-p)^2+4(m+4p)(4+pm)\geq 0$$
for any choice of $p$.
$$(9+16m)p^2+(4m^2-46)p+9+16m \geq0$$
Then we have to analyze that new quadratic equation. Once the above expression must be always non negative, then.
$$\Delta'=(4m^2-46)^2-4(9+16m)^2=(m^2-8m-16)(m^2+8m-7)\leq 0$$
Solving the above inequality we have the values of $m$.
