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 $13$.