The sum of the squares of two numbers is 233 and

Let the numbers be integers. One of the numbers be $x$. So, the other will be $(2 x-3)$.

Then according to question,

$x^{2}+(2 x-3)^{2}=233$

$x^{2}+4 x^{2}-12 x+9=233$

$5 x^{2}-12 x+9-233=0$

$5 x^{2}-12 x-224=0$

$5 x^{2}-40 x+28 x-224=0$

$5 x(x-8)+28(x-8)=0$

$(x-8)(5 x+28)=0$

$(x-8)=0$

$x=8$

Or

$(5 x+28)=0$

$x=\frac{-28}{5}$

Since, we have assumed the numbers to be integers, so x cannot be a rational number/fraction.

Therefore, for x = 8

Other number =

$(2 x-3)=2 \times 8-3$

$=16-3$

$=13$

Thus, whole numbers be 8,13 .