Find the value of $x$, for which the function $h(x) = \frac 1{(x-5)^2+4(x-5)+4}$ is undefined Picture of problem
I'm having an issue with a simple SAT question. If you can't see from the picture, I'm being asked to find the value of $x$, for which the function:
$$h(x) = \frac 1{(x-5)^2+4(x-5)+4}$$
is undefined. 
I figured that for it to be undefined, it should be equal to 1/0. This therefore means that $(x-5)^2+4(x-5)+4 = 0$. I further simplified it into 
$$(x-5)((x-5)+4) = 0$$
However, this gives me either $x = 1$ or $x = -3$, both of which are false. The answer is $3$. Why do I get $-3$, when the answer is $3$?

The denominator has to be zero for the chosen $x$ values of course;
$(x-5)^2+4(x-5)+4=0$
We can name $(x-5)=a$ to make things easier.
$a^2+4a+4=0$ 
That is a perfect square;
$(a+2)^2=0$
$a=-2$ so $x-5=-2 \rightarrow x=3$
