The sum of a number and its square is 63/4,

Let first numbers be x

Then according to question

$x+x^{2}=\frac{63}{4}$

Let $x=y^{2}$ then

$4\left(x+x^{2}\right)=63$

$4 x^{2}+4 x-63=0$

$4 x^{2}+18 x-14 x-63=0$

$2 x(2 x+9)-7(2 x+9)=0$

$(2 x+9)(2 x-7)=0$

$(2 x+9)=0$

$x=-\frac{9}{2}$

Or

$(2 x-7)=0$

$x=\frac{7}{2}$

Thus, the required number be $\frac{7}{2}, \frac{-9}{2}$