jmc

Expanding the product on the right, we have $x^2+4x+3=-(x^2+8x+15),$ so $x^2+4x+3+(x^2+8x+15)=0$. Simplifying the left side gives $2x^2+12x+18=0.$ Dividing by 2, we have $x^2+6x+9=0$, so $(x+3)(x+3)=0.$ The only solution for $x$ is $\boxed{-3}.$