x4-x=10.
Let f(x)=x4-x-10, f'(x)=4x3-1.
Use iteration formula: xn+1=xn-f(xn)/f'(xn).
We need to find a suitable value for x0.
f(1)=-10 and f(2)=2 so there's a positive root between x=1 and x=2, therefore let x0=1.
x1=1-(-10)/3=13/3, x2=3.29..., x3=2.098..., x4=1.8956..., x5=1.895688..., x6=1.85558..., x7=1.855584529, x8=x7 so x=1.855584529 is an approximate positive zero. A more accurate value is 1.85558452864.
We could have used x0=2 and reached the same accuracy at x4.