Solve for $x$: $\sum_{r=0}^5 {5\choose r} (-1)^rx^{5-r}3^r = 32$

Question

Solve for $x$: $$\sum_{r=0}^5 {5\choose r} (-1)^rx^{5-r}3^r = 32$$

Looks like binomial theorem. So this would simplify to $(-x+3)^5=32$, and solving gives $x=2$. Is this correct?