Since the ordinates are same

-x^2-2x+6 = x^2+2x+1

0 = x^2+2x+1 +x^2 +2x-6

0 = 2x^2+4x-5

x = (-4+or-(4^2+4*2*5)^(1/2)}/2*2

= (-4+or- sqrt(56)}/4

= (-2+sqrt18)/2 or (-2-sqrt18)/2.

The corresponding y could be calculated by any one of the equations. So the equation has two solutions with with 2 points of intersections.

To identify the number of solutions of the system, we'll have to form the identity:

x^2+x+1=-x^2-2x+6

We'll move all the terms to one side and we'll obtain a quadratic equation:

x^2+x+1+x^2+2x-6=0

2x^2+3x-5=0

We'll apply the quadratic formula:

x1=[-3+sqrt(9+40)]/4

x1=(7-3)/4

x1=1

x2=(-7-3)/4

x2=-10/4

x2=-5/2

For finding y1, we'll substitute x1 into one of the 2 equations, for example x^2+x+1=y

x1=1

y1=1+1+1

y1=3

x2=-5/2

y2=(-5/2)^2+(-5/2)+1

y2=25/4 - 5/2 + 1

y2=(25-10+4)/4

y2=19/4

**So, the solutions of the system are 2 and they are:**

**{(1,3) ; (-5/2,19/4)}**