solving rectangle Diagonal of a rectangle is $13$ cm. If we extend the length of the rectangle for $4$ cm and width for $7$ cm, then diagonal will be longer for $7$ cm as well. Find sides (length and width) of the rectangle.
So $d^2=a^2+b^2 \implies b^2=d^2-a^2 \implies b^2=13^2-a^2 \implies b^2=169-a^2$ and $a^2+b^2=169$
\begin{align*}
d'^2 & =a'^2+b'^2\\
(d+7)^2 & =(a+4)^2+(b+7)^2\\
20^2 & = a^2+8a+16+b^2+14b+49\\
400 & = 8a+16+14b+49+169\\
\frac{166}{2} & = \frac{8a+14b}{2}\\
83 & =4a+7b
\end{align*}
... 
That's what I've done. What should I do next?

You have
$$
a^2+b^2=169
$$
$$
4a+7b=83 \Longleftrightarrow a=\frac{83-7b}{4}
$$
After substitution:
$$
\frac{(83-7b)^2}{16}+b^2=169
$$
$$
\frac{49b^2}{16}-\frac{1162b}{16}+\frac{6889}{16}-169=0
$$
Finally, use 'abc-formula' with a = 49/16, b = 1162/16 and c = 6889/16-169.
