Intermediate Algebra Quadratic Inequalities 
For what values of $a$ does the equation $$(a^{2}+2a)x^{2}+(3a)x+1=0$$ yield no real solutions $x$? Express your answer in interval notation.

What I did so far:
The only time a quadratic equation has no real solutions is when the discriminant is less than $0$. In this case, the discriminant equals $9a^2-4a^2-8a$, and that is less than $0$. Simplifying, I got $5a^2-8a<0$, so $a(5a-8)<0$. I have $2$ cases: $$a<0$$ $$5a-8>0$$ and $$a>0$$ $$5a-8<0.$$ The first case is impossible because $a$ must be positive and negative at the same time. The second case says that $a>0$ and $a<\frac{8}{5}$, which can be written as $(0,\frac{8}{5})$ in interval notation, but that answer wasn't correct.
Could I get a little help? Thanks!
 A: There is one subtle special case you're overlooking.
You assume you're dealing with a quadratic, but what if what you're handling is not a quadratic at all, i.e. is something of lesser degree?
Well, then there are two further cases:

*

*$f(x) = (a^2+2a)x^2+(3a)x+1$ is linear (and for simplicity, is not constant)

*$f$ is constant

Can either hold, with having no real roots?
In the former case, no. (Note that assuming $f$ is not constant is valid, since otherwise just $f$ being linear would subsume the constant case, and a distinction needs to be made.) Any line with non-zero slope will cross the $x$-axis.
Can $f$ be constant? If $f$ is constant, then the coefficients of $x^2,x$ must be zero. In that event, we would have $f(x) = 1$, which definitely has no real roots.
But the question remains, can this occur, and for which $a$? Well, the equations
$$a^2 + 2a = 0 \qquad 3a = 0$$
must be satisfied. It is easy to see that this requires $a=0$.

Hence, I would imagine the intended answer is actually $[0,8/5)$.
Otherwise, your logic holds up. (You can also play with this Desmos demo to see how the function $f$ behaves for varying values of $a$.)
