Parabolas

Set the equations equal

8x^2 + 6x + 1 =  2x^2 - 7x + 1          rearrange as

6x^2 + 13x   =  0       factor

x (6x  - 13)  = 0

The  x coordinates   are

x = 0                        and        6x - 13 = 0

                                                6x = 13

                                                   x  =13/6

When x = 0  y  =1

When x = 13/6, y=  2(13/6)^2 - 7(13/6) + 1   =  -43/9

The points of intersection are 

(0 , 1) , (13/6 , -43/9 )

Set equal: $2x^2-7x+1 =8x^2+6x+1$

Simplify: $6x^2 + 13x = 0$

Factor: $x(6x + 13) = 0$

If $6x + 13 = 0$, $x = -{13 \over 6}$, else $x = 0$

Substituting this into the quadratic, we find the points of intersection are $\color{brown}\boxed{(-{13 \over 6},{ 43 \over 9}) \ \text{and} \ ({0, 1})}$