Is there a way to find expected value of equation?

Since $X$ is binomial you also know that $Var(X)=np(1-p)$. So, use that (this holds in general) $$Var(X)=E[X^2]-E[X]^2\implies E[X^2]=Var(X)+E[X]^2$$ and linearity of expectation to obtain that
\begin{align}E[4+3X^2]&=E[4]+E[3X^2]=4+3E[X^2]=4+3(Var(X)+E[X]^2)\\[0.2cm]&=4+3(np(1-p)+(np)^2)\end{align}