Answer
$(-5,1)$, $(-\frac{1}{3},-\frac{5}{9})$
Work Step by Step
1. Let $f(x)=g(x)$, we have $x^2+5x+1=-2x^2-11x-4\Longrightarrow 3x^2+16x+5=0\Longrightarrow (3x+1)(x+5)=0\Longrightarrow x=-5,-\frac{1}{3}$
2. For $x=-5$, we have $f(-5)=(-5)^2+5(-5)+1=1$, giving point $(-5,1)$
3. For $x=-\frac{1}{3}$, we have $f(-\frac{1}{3})=(-\frac{1}{3})^2+5(-\frac{1}{3})+1=-\frac{5}{9}$, giving point $(-\frac{1}{3},-\frac{5}{9})$
