jmc

First, set the two equations equal to each other to get $2x^2-7x+1=8x^2+5x+1$. Combine like terms to get $6x^2+12x=0$. Then, we can divide by $6$ to get $x^2+2x=0$. To complete the square, we need to add ${\left(\frac{2}{2}\right)}^2=1$ to both sides, giving $(x+1{)}^2=1$.

So we have $x+1=\pm 1$. Solving for $x$ gives us $x=-2$ or $0$. Using these in our original parabolas, we find the points of intersection to be $\boxed{(-2,23)}$ and $\boxed{(0,1)}$.