Algebra question

Let $x$ be the length of the sides of the smaller square, and $y$ be the length of the sides of the larger square. Then 
\[(1)        y=3+x.\]
The sum of the areas of the squares is 
\[(2)        x^2+y^2=185.\]
Substituting (1) in (2) we get, 
\[x^2+(x+3)^2=185  \Rightarrow x^2+x^2+6x+9=185\]
\[\Rightarrow   2x^2+6x-176=0\]
\[\Rightarrow x=\frac{-6 \pm\sqrt{36-4(2)(-176)}}{2(2)}=\frac{-6 \pm\sqrt{1444}}{4}=\frac{-6 \pm 38}{4}=-11, 8.\]
$x=-11$ is not acceptable, so $x=8$ is the only solution. Therefore the length of the sides of the larger square is $y=3+8=11$.