jmc

Since the coefficients are real, the other roots must be $-2+3i.$ Thus, the quadratic is a constant multiple of $$(x+2+3i)(x+2-3i)=(x+2{)}^2-(3i{)}^2=(x+2{)}^2+9=x^2+4x+13.$$We want the coefficient of $x$ to be $-4,$ so we simply multiply this quadratic by $-1,$ to get $\boxed{-x^2-4x-13}.$