Note that $(x-2)(x+7)=x^2 + 5x - 14$ and $(x+1)(x+4)=x^2+5x+4$ only differ in constant terms. Let $$y=\frac{(x-2)(x+7) + (x+1)(x+4)}{2}= x^2 + 5x -5$$ Then \begin{align} (x-2)(x+1)(x+4)(x+7)&=19\\ \Big((x-2)(x+7)\Big)\Big((x+1)(x+4)\Big)&=19\\ (y-9)(y+9)&=19\\ y_{1,2}&=\pm 10 \end{align} When $y=10$, $x^2-5x-5=10 \implies x_{1,2}=\boxed{\frac{-5\pm\sqrt{85}}{2}}$. When $y=-10$, $x^2-5x-5=-10 \implies x_{3,4}=\boxed{\frac{-5\pm\sqrt{5}}{2}}$.