Give exact and approximate solutions to three decimal places x^2+2x+1=36

Please help! Thanks!

x^2 + 2x + 1 = 36
x^2 + 2x - 35 = 0
(x-5)(x+7) = 0
x = 5 or -7

Exact solution: x = 5, and x = -7. x^2 + 2x - 35 = 0, (x-5)(x+7) = 0

Use Newton's Method.

f (x) = x^2 + 2x - 35.
f ' (x) = 2x + 2
Let Xo = 0

X1 = Xo - f(Xo) / f ' (Xo)
X2 = X1 - f(X1) / f ' (X1)
etc

x^2+2x+1=36
(x+1)² = 36
x = -1±6
x = -7, 5