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

Not only the equation in $y$ must have real roots, but these roots have to be $\le 1$, since it's asked that the equation in $x$ has non-positive roots, so that $y=2^x\le 2^0=1$.
To test whether the roots are less than $1$, we can place $1$ w.r.t. the roots $y_1,y_2$  of $\;p(y)=y^2-(a-3)y+a+4$. The standard method is to determine the sign of $p(1)=8>0$. Therefore $1$ does not separate the roots of $p(y)$, and  $1$ is either less than or greater than both of them. Furthermore, the half-sum of the roots is $\;\frac{a-3}2$ by Vieta's relations, so that
$$y_1, y_2 \le 1\iff \frac{y_1+y_2}2=\frac{a-3}2 \le 1\iff a\le 5. $$
So the solution is
$$a\in \Bigl\{\Bigl(-\infty,5-4\sqrt{2}\Bigr]\cup \Bigl[5+4\sqrt{2},\infty\Bigr)\Bigr\}\cap (-\infty,5]=\Bigl(-\infty,5-4\sqrt{2}\Bigr].$$
