1/ 3(x^3) + 1/2(x^2) + 1 = 0, x1 = −3 newtons way to the 4th decimal

1/ 3(x^3) + 1/2(x^2) + 1 = 0

times 6
2x^3 + 3x^2 + 6 = 0
let y = 2x^3 + 3x^2 + 6

dy/dx = 6x^2 + 6x

according to Newton's method
newx = x - y/y'
= x - (2x^3 + 3x^2 + 6 )/(6x^2 + 6x)
= (6x^3 + 6x^2 - 2x^3 - 3x^2 - 6)/(6x^2 + 6x)
= (4x^3 + 3x^2 - 6)/(6x^2 + 6x)

using my calculator, x = -3
newx = -2.41666...
x = -2.41666.. , newx = -2.1875......
x = -2.1875.. , newx = -2.150313..
x = -2.150313... newx = -2.1493768..
x = -2.1493768 newx = -2.1493762

x = -2.1494 correct to 4 decimal places