What is the intersepting point of y = 2x^2 - 7 and the line y= -2x + 1

To determine the intercepting point of the line and parabola,
we'll have to solve the system:

2x^2 - y =
7 (1)

2x + y = 1 (2)

We'll calculate x
form the 2nd equation:

2x = 1 - y

x =
(1-y)/2

We'll substitute x in
(1):

2*(1-y)^2/4 - y = 7

(1-y)^2/2 - y
= 7

We'll multiply by 2 both
sides:

(1-y)^2 - 2y - 14 = 0

We'll
expand the square:

1 - 2y + y^2 - 2y - 14 =
0

We'll combine like terms and we'll re-arrange the
terms:

y^2 - 4y - 13 = 0

We'll apply
the quadratic:

y1=[4+sqrt(16+52)]/2

y1
= 2 + sqrt17

y2 = 2 - sqrt17

x1 = 
(1-y1)/2

x1 = (-1-sqrt17)/2

x2 =
(-1+sqrt17)/2

The intercepting points
are:

((-1-sqrt17)/2 , 2 + sqrt17) and
((-1+sqrt17)/2 , 2 - sqrt17).