(x-2)^2+(y-3)^2=6 and y=-2x^2+14x-1 are the two equations, but I dont know how to start finding all the point of intersections. Trying substitution got me a quartic equation

x^2 - 4x + y^2 - 6y = -7 [1]

(-2x^2 + 14x - 1) = y [2]

x^2 - 4x + (-2x^2 + 14x - 1)^2 - 6(-2x^2 + 14x - 1) = -7

4x^4-56x^3+213x^2-116x+14=0