Quadratic

ax^2 + 18x + 9   =        (3x+3)^2          when a = 9

The expression $ax^2+15x+4+3x+5$ simplifies to $ax^2+18x+9$.  Since it's the square of a binomial, it must equal $(cx+d)^2$ for some constants $c$ and $d$.  Now $(cx+d)^2= c^2x^2+2cdx+d^2$ and we can equate its coefficients to that of $ax^2+18x+9$ like so:
$$\begin{eqnarray*} a &=& c^2 \\ 2cd &=& 18 \\ d^2 &=& 9 \end{eqnarray*}$$
We solve this system of 3 equations in 3 unknowns $a$, $c$, and $d$.   Note that there are two solutions but both solutions have the same value of $a$.