exponential equation has non positive roots 
Find real values of $a$ for which the equation
$4^x-(a-3)\cdot 2^x+(a+4)=0$
has non positive roots

Try: Let $2^x=y\in (0,1].$ Then equation convert into
$y^2-(a-3)y+(a+4)=0$
For real roots its discriminant $\geq 0$
So $$(a-3)^2-4(a+4)\geq 0$$
$$a^2-10a-7\geq 0$$
$$a\in \bigg(-\infty,5-4\sqrt{2}\bigg]\cup \bigg[5+4\sqrt{2},\infty\bigg).$$
but answer is different from that , i did not know where i
am missing. could some help me to solve it. thanks

You want $f(y) = y^2 - (a-3) y + (a+4)$ to have at least one root in $(0,1]$.
Note that $f(0) = a+4$ and $f(1) = 8$, while the minimum value of $f(y)$, occurring at 
$y = (a-3)/2$, is $(-a^2 + 10 a + 7)/4$.  There are several cases to consider.


*

*If $a < -4$, $f(0) < 0 < f(1)$ so there is a root in $(0,1]$.

*For $a = -4$, $f(y) = y^2 + 7 y$ so the roots $0$ and $-7$ are not in $(0,1]$. 

*For $a > -4$, $f(0) > 0$ and $f(1) > 0$ so the only way to have a root in $(0,1]$ would be if the minimum occurs between $0$ and $1$ and the minimum value $\le 0$.  The
minimum is at $(a-3)/2$, so this requires $a \in (3,5]$.  But for $a$ in this interval,
$-a^2 + 10 a + 7 > 0$.


So the answer is $a < -4$.
Alternative method:
For given $y$, $f(y) = 0$ for $a = A(y) = 4 + y + 8/(y-1)$.  Thus you want the set of values $A(y)$ for $0 < y < 1$ (of course $A(1)$ is undefined).
$A'(y) =  \dfrac{y^2 - y - 7}{(y-1)^2} < 0$ in this interval, so $A$ is a decreasing function of $y$, with $A(y) \to -\infty$ as $y \to 1-$.  Thus the answer is $(-\infty, A(0)) = (-\infty, -4)$.  
