Coefficent of an equation I am suppose to find the coefficent of x^2 in this equation

after doing the calculation I ended up with this

but that is the wrong answer what is that I am missing?

You have $f(x)=x^2+bx+c$, and you want $f\big(f(x)\big)$. As you say, this is $$\left(x^2+bx+c\right)^2+b\left(x^2+bx+c\right)+c\;.\tag{1}$$
Now
$$\begin{align*}
\left(x^2+bx+c\right)^2=x^4+2bx^3+\left(b^2+2c\right)x^2+2bcx+c^2\;,
\end{align*}$$
so $(1)$ becomes
$$x^4+2bx^3+\left(b^2+2c+b\right)x^2+\left(2bc+b^2\right)x+c^2+bc+c\;,$$
in which the coefficient of $x^2$ is $b^2+2c+b$. If $b=-23$ and $c=15$, this is
$$(-23)^2+2(15)-23=529+30-23=536\;.$$
I did it this way to follow (roughly) your approach. Note, however, that you don’t actually have to multiply everything out: you just need the $x^2$ term from $\left(x^2+bx+c\right)^2$ and the $x^2$ term from $b\left(x^2+bx+c\right)$. The latter is clearly $bx^2$. The former can only come from the terms $\left(x^2\right)(c)$, $(bx)(bx)$, and $(c)\left(x^2\right)$ in $\left(x^2+bx+c\right)^2$, so the $x^2$ term here is $\left(b^2+2c\right)x^2$. Now just add the coefficients to get $b^2+2c+b$, and do the arithmetic as before.
