Solution:-
let us assume the two numbers be y, $2y – 3$
As per the condition given in the question, \[{{y}^{2}}~+\text{ }{{\left( 2y-3 \right)}^{2}}~=\text{ }233\]
Then,
\[{{y}^{2}}~+\text{ }{{\left( 2y-3 \right)}^{2}}~=\text{ }233\]
We know that,
\[{{\left( a-b \right)}^{2}}~=\text{ }{{a}^{2}}-2ab\text{ }+\text{ }{{b}^{2}}\]
\[{{y}^{2}}~+\text{ }4{{y}^{2}}-12y\text{ }+\text{ }9\text{ }=\text{ }233\]
\[5{{y}^{2}}-12y\text{ }+\text{ }9\text{ }=\text{ }233\]
By transposing we get,
\[5{{y}^{2}}-12y\text{ }+\text{ }9-233=\text{ }0\]
\[5{{y}^{2}}-12y-224\text{ }=\text{ }0\]
\[5{{y}^{2}}-40y\text{ }+\text{ }28y-224\text{ }=\text{ }0\]
Take out common in each terms,
$5y(y – 8) + 28(y – 8) = 0$
$(y – 8) (5y + 28) = 0$
Equate both to zero,
$y – 8 = 0$
$5y + 28 = 0$
$y = 8$
$5y = – 28$
$y = 8$
$y = – 28/5$
So, $y = 8$ … [because 8 is a natural number]
Then,
$y = 8$
$2y – 3 = 2(8) – 3$
$= 16 – 3$
$= 13$
Therefore, the $2$ numbers are $8$ and $13$.