geometry - Given the lenght of the hypotenus and the sum of the legs find the area of right triangle

Given: right-angled $\triangle ABC$, the length of the hypotenuse $c = 5$ and the sum of the legs $a+b = 6$. Find the area of the triangle.
I tried creating the following system $a+b = 6; a^2+b^2 =25$ however it got me nowhere, I get a really ugly quadratic equation .

Answer

We have
\begin{align*}
\text{Area }&=\frac12ab\sin \angle C\\[3pt]
&=\frac12ab\sin\frac{\pi}2\\[3pt]
&=\frac12ab\\[3pt]
&=\frac14\left[(a+b)^2-(a^2+b^2)\right]\\[3pt]
&=\frac14(36-25)\\[3pt]

&=\frac{11}4
\end{align*}