To determine the points that lie on both the curves y=x^2+x+1 and y=-x^2-2x+6, we have to equate the two. Doing this gives

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

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

=> 2x^2 +...

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To determine the points that lie on both the curves y=x^2+x+1 and y=-x^2-2x+6, we have to equate the two. Doing this gives

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

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

=> 2x^2 + 3x - 5 = 0

=> 2x^2 + 5x - 2x - 5 =0

=> x ( 2x + 5) - 1( 2x + 5) = 0

=> ( x -1)( 2x + 5) = 0

This gives x = 1 and x = -5/2

At x = 1, y = x^2 + x + 1 = 1 + 1 + 1 = 3

At x = -5/2 , y = (-5/2)^2 - 5/2 + 1 = 19/4

**Therefore the points that lie on both the curves are ( 1, 3) and ( -5/2 , 19/4).**