How is discriminant related to real $x$? Question in my text book
Solve for range of the function, $$y=\frac{x^2+4x-1}{3x^2+12x +20}$$
Text book says, cross multiply and express the obtained equation as a quadratic equation in $x$
So I get $ (3y-1)x^2 + (12y -4)x + (20y-1) = 0$
Now it says find discriminant $D$, so we have,
$$ D = -4 ( 3y-1)( 8y+3)$$
Now it says, set $D≥0$ as $x$ is real.
Wait what?
Isn't $x \in \mathbb{R}$ the domain for a quadratic function? Meaning "$x$" is always real? What does a discriminant got anything to do with $x$ being real, when all discriminant tells us is whether or not the ROOTS are real? Help please.
To be more specific about my doubt, here's an edit.
EDIT : I'm confused, setting discriminant $≥0$ would tell whether or not roots are real, meaning whether the graph of the quadratic function cuts/touches X axis. Now tell me what does this got anything to do with range? As far as I know, quadratic fucntions that don't have real roots are also continous throughout the X axis, meaning there SHOULD always be a corresponding $y-value$ 

$$\frac{x^2+4x-1}{3x^2+12x+20}=y$$
After cross multiplying, we obtain 
$$(3y-1)x^2+(12y-4)x+(20y+1)=0$$
If the discriminant is nonnegative, then we can find real $x$ as solution to the quadratic equation, and hence given a particular value of $y$ that makes the discriminant nonnegative, we can find $x$ that satisfy the original equation.
However, if the discriminant is negative, then $x$ that satisfies the quadratic equation is not real.
