Answer
$x^2+6x+4$
Work Step by Step
Write the quadratic as the product of two linear factors and the leading coefficient: $a(x-r_1)(x-r_2)=0$, where $r_1$ and $r_2$ are the solutions.
$a[x-(-3+\sqrt 5)][x-(-3-\sqrt 5)]=0~~$ (Make $a=1$ since it can be any real number except $0$)
$(x+3-\sqrt 5)(x+3+\sqrt 5)=0$
$(x+3)^2-(\sqrt 5)^2=0$
$x^2+6x+9-5=0$
$x^2+6x+4$ (general form)
