Find the real parameter so as the equation has no real solutions.... My question is: 
For which values of parameter $a\in \mathbb{R}$ the following equation
$$25^x+(a-4) \,5^x-2a^2+a+3=0$$
has no real solutions?

My idea is:
First of all we should transform the equation above in a quadratic equation but noting $5^x=t$ where $t\gt0$. So first of all I think that a quadratic equation has no real solutions when $D\lt 0$, but the second case , I think is when $D\gt0$, but the solutions are negative. But although it seems like logical, I still do not get the right, convenient solutions. 
I hope you help me solve this problem. Thank you very much!

\begin{align}
25^x+(a-4) \,5^x-2a^2+a+3&=5^{2x}+(a-4) \,5^x-2a^2+a+3\\
&=(5^x)^2+(a-4) \,5^x-2a^2+a+3
\end{align}
Let $y=5^x$, then
\begin{align}
(5^x)^2+(a-4) \,5^x-2a^2+a+3=y^2+(a-4) \,y+3+a-2a^2=0
\end{align}
The quadratic equation has no real roots iff its discriminant less than zero. Hence
\begin{align}
D&=0\\
(a-4)^2-4\cdot1\cdot(3+a-2a^2)&=0\\
a^2-8a+16+8a^2-4a-12&=0\\
9a^2-12a+4&=0\\
(3a-2)^2&=0\\
a&=\frac{2}{3}
\end{align}
In this case, its discriminant will be negative if $a<\cfrac{2}{3}$.
