I've noticed it would have a parabola shape and subtracting 3 creates a y-axis crossing point at y=-3.
You don't. It's not true. Or are you asking about real roots?
Is that really the problem? so so that there are four roots, the four fourth roots of -3/2, none of which are real.
If the equation is , we can let so we are dealing with the quadratic equation . By the quadratic formula,
which has the two roots y= 6/8= 3/4 and y= -8/8= -1.
Now, If , then , two real roots. If , then , two imaginary roots.
The crucial point was that, because this had only even factors, we could reduce to a second order equation. Positive roots of that equation give real roots of the original equation.