Finding the range of a Quadratic divided by a Quadratic I need to find the range of a function:
$$f(x) = \frac{e^{2x} - e^x + 1}{e^{2x} + 3e^x - 7} $$
Substituting $e^x = t $ and equating $f(x)$ to $y$, I get
$$(y-1)t^2 + (3y+1)t - (7y+1) = 0$$
Since $t$ must be real and positive, I get the range of y to be $(-\infty, -\frac{1}{3})\bigcup(1, \infty) $.
Is this correct?

Substitute $f(x)$ with $y$: $y=\frac{e^{2x}-e^x+1}{e^{2x}+3e^x-7}$.
Rewrite in terms of $e^x$:
$$e^{2x}y+3e^xy-7y=e^{2x}-e^x+1$$
$$(y-1)(e^x)^2+(3y+1)e^x-(7y+1)=0$$
This equation has roots (with respect to $x$), when there are roots with respect to $e^x$ (discriminant is non-negative) and at least one root is positive.
$D=(3y+1)^2-4(y-1)(-7y-1)=37y^2-18y-3\ge0$ or $y\in\left(-\infty, \frac{9-8\sqrt{3}}{37}\right]\cup\left[\frac{9+8\sqrt{3}}{37}, \infty\right)$
In this case $e^x=\frac{-(3y-1)\pm\sqrt{37y^2-18y-3}}{2(y-1)}$
