Find the values of $b$ for which the equation $2\log_{\frac{1}{25}}(bx+28)=-\log_5(12-4x-x^2)$ has only one solution Find the values of 'b' for which the equation 
$$2\log_{\frac{1}{25}}(bx+28)=-\log_5(12-4x-x^2)$$ has only one solution.
=$$-2/2\log_{5}(bx+28)=-\log_5(12-4x-x^2)$$
My try:
After removing the logarithmic terms I get the quadratic $x^2+x(b+4)+16=0$
Putting discriminant equal to $0$ I get $b={4,-12}$
But $-12$ cannot be a solution as it makes $12-4x-x^2$ negative so I get $b=4$ as the only solution.
But the answer given is $(-\infty,-14]\cup{4}\cup[14/3,\infty)$.I've no idea how.Help me please.

You have
$$x^2+(4+b)x+16=0\tag1$$
This is correct.
However, note that when we solve
$$2\log_{\frac{1}{25}}(bx+28)=-\log_5(12-4x-x^2)$$
we have to have
$$bx+28\gt 0\quad\text{and}\quad 12-4x-x^2\gt 0,$$
i.e.
$$bx\gt -28\quad\text{and}\quad -6\lt x\lt 2\tag2$$
Now, from $(1)$, we have to have $(4+b)^2-4\cdot 16\geqslant 0\iff b\leqslant -12\quad\text{or}\quad b\geqslant 4$.
Case 1 : $b\lt -14$
$$(2)\iff -6\lt x\lt -\frac{28}{b}$$
Let $f(x)=x^2+(4+b)x+16$. Then, since the equation has only one solution, we have to have
$$f(-6)f\left(-\frac{28}{b}\right)\lt 0\iff b\lt -14$$
So, in this case, $b\lt -14$.
Case 2 : $-14\leqslant b\leqslant -12$ or $4\leqslant b\lt \frac{14}{3}$
$$(2)\iff -6\lt x\lt 2$$
$b=4$ is sufficient, and $b=-12$ is not sufficient. For $b\not=4,-12$,
$$f(-6)f(2)\lt 0\iff b\lt -14\quad\text{or}\quad b\gt \frac{14}{3}$$
So, in this case, $b=4$.
Case 3 : $b\geqslant \frac{14}{3}$
$$(2)\iff -\frac{28}{b}\lt x\lt 2$$
$b=\frac{14}{3}$ is sufficient. For $b\gt\frac{14}{3}$,
$$f\left(-\frac{28}{b}\right)f(2)\lt 0\iff b\gt \frac{14}{3}$$
So, in this case, $b\geqslant 14/3$.
Therefore, the answer is $$\color{red}{(-\infty,-14)\cup{4}\cup\bigg[\frac{14}{3},\infty\bigg)}$$
