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!
 A: For real solution the discriminant of $$(5^x)^2+5^x(a-4)+3+a-2a^2=0$$ must be $\ge0$
i.e., $$(a-4)^2-4\cdot1\cdot(3+a-2a^2)=9a^2-12a+4=(3a-2)^2$$ needs to be $$\ge0$$ which holds true for all real $a$
So, we have $$5^x=\frac{-(a-4)\pm(3a-2)}2$$
Now for real $\displaystyle x, 5^x>0$
A: \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 $a<\cfrac{2}{3}$.
A: Another answer got as far as
\[5^x=\frac{-(a-4)\pm(3a-2)}2\]
observed that $5^x$ must be positive, but didn't comment on what this implied about $a$.
Well, let's take the $\pm$ to be $+$. Then
\begin{aligned}
&\frac{-(a-4) + (3a-2)}2 > 0 \\
\implies&2a+2>0\\
\implies&a>-1,\end{aligned}
whereas, if that $\pm$ is $-$, then:
\begin{aligned}
&\frac{-(a-4) - (3a-2)}2 > 0 \\
\implies&-4a+6>0\\
\implies&a<\textstyle\frac{2}{3}.\end{aligned}
Only one of these conditions need to be satisfied for a solution to exist, so there is always a solution for every value of $a$ (but, unusually for a quadratic, there is usually only one).
