Problem with Newton's Method I am trying to solve this problem:

Use Newton's method to find the intersection points of the two circles defined by
$$x^2 + y^2 = 2, \quad (x-1)^2 + (y+1.5)^2 = 1.$$

I used this code in Maple:
Newton(4+(-y^2+2)*(2*y+3)^2-4*(-y^2+2), y = .5, output = sequence);

and this gave me:
.5, -1.062500000, -.7532718336, -1.001899321, -.5138663374,
-.7942990696, -1.069866630, -.7693943666, -1.025772773,
-.6418803790, -.8854387837

From this I do not understand what the solution is, and also how can I choose my initial value more accurately?

First I couldn't understand why you used that particular formula. the formula should be:
$$(x-1)^2 + (y + \frac 32)^2 = 1$$
$$x^2 - 2x + 1 + y^2 + 3y + \frac 94 = 1$$
$$(x^2 + y^2) - 2x + 3y + \frac 94 = 0$$
$$2 - 2x + 3y + \frac94 = 0$$
$$3y - 2x + \frac{17}{4} = 0$$
$$y = \frac{2x - \frac{17}{4}}{3}$$
Now substituting we have:
$$x^2 + \left(\frac{2x - \frac{17}{4}}{3}\right)^2 - 2 = 0$$
$$x^2 + \frac{4x^2 - 17x + \frac{289}{16}}{9} - 2 = 0$$
$$9x^2 + 4x^2 - 17x + \frac{289}{16} - 18 = 0$$
$$144x^2 + 64x^2 - 272x + 289 - 288 = 0$$
$$208x^2 - 272x + 1 = 0$$
Now it should be easy to find the solutions using Newton's method.
Also Newton's method may fail for numbers of reasons. First is "bad" starting point. For some starting points the value of x doesn't converges to the right value, but diverges
For more info you can read here
