Hello, Lizarrdo!
$\displaystyle \text}Find the possible values of }x.$
Code:C * *| * * | * x * |h * * | * * | * A * * * * * * * B x-5 D x-3
Note that $\displaystyle x > 5.$
We have right triangle $\displaystyle ABC,\:\angle C = 90^o.$
The altitude to the hypotenuse is: .$\displaystyle h = CD.$
It divides the hypotenuse into two segments: .$\displaystyle AD = x-5,\;DB = x-3.$
Fact: The altitude to the hypotenuse is the geometric mean
. . . . . . . . . of the segments of the hypotenuse.
Hence, we have: .$\displaystyle h^2 \:=\:(x-3)(x-5)$ .[1]
In right triangle $\displaystyle CDB\!:\;h^2 + (x-3)^2 \:=\:x^2 \quad\Rightarrow\quad h^2 \:=\:x^2-(x-3)^2$ .[2]
Equate [1] and [2]: .$\displaystyle (x-5)(x-3) \:=\:x^2 - (x-3)^2$
We have: .$\displaystyle x^2 -14x + 24 \:=\:0 \quad\Rightarrow\quad (x-2)(x-12) \:=\:0$
. . Therefore: .$\displaystyle x \,=\,12$