Triangle $\triangle ABC$ is a right triangle with $\measuredangle ACB=90^\circ$. Let $AB=c$ and the radius of the inscribed circle be $r$. Find the catheti and the area of the triangle $\triangle ABC$.
Let $P,N$ and $M$ be the tangent points with $AB,BC$ and $CA$, respectively. The quadrilateral $MINC$ is a square. Therefore, $MI=IN=NC=CM=r.$ We have $AM=AP=c_1$ and $BN=BP=c_2$ as tangent segments. Now we have: $$\begin{cases} (c_1+r)^2+(c_2+r)^2=(c_1+c_2)^2 \\ c_1+c_2=c \Rightarrow c_2=c-c_1 \end{cases}.$$ After simplifying the first equation I got $r^2+c_1r+c_2r=c_1c_2$ and now let us plug $c_2=c-c_1$. We get the quadratic equation $c_1^2-cc_1+r^2+cr=0$ which has roots $c_1=\dfrac{c\pm\sqrt{c^2-4r^2-4cr}}{2}$. I am not sure I understand what to do next and why we got two expressions for the same segment. What does that mean? If use use both the results in $c_2=c-c_1=\dfrac{c\pm\sqrt{c^2-4r^2-4cr}}{2}$. What does mean?
You got two solutions, because the triangle has two catheti. It is not unusual that the equation (if correctly composed) appears to be smarter than it was supposed by its construction.
In the considered case the equation gives the solutions for both catheti, and this is in fact inavoidable. Indeed you system of equations is symmetric with respect to $c_1$ and $c_2$. Therefore it must have in general case at least two solutions: $(a,b)$ and $(b,a)$. Hence, it is not at all surprising that the equation for $c_i$ appears to be quadratic. And since the equation is quadratic, the solution of the problem is unique up to a congruence relation between the resulting triangles. Which one of the two solutions you name $c_1$ and which $c_2$ plays no role.
Your next step is to compute the catheti using the relations $a=c_1+r$ and $b=c_2+r $. The rest is trivial.