$Algebra question$

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$Algebra question$

The length of each side of a square is $3 in.$ more than the length of each side of a smaller square. The sum of the areas of the squares is $185$ squared in. Find the lengths of the sides of the two squares.

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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 $y=3+8=11$.

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