Question 821367
Assuming you want an approximate real solution we can use Newton's method. Writing f(x)=x^3+10x-80, and f'(x)=3x^2+10 and writing as a sequence we can see:

{{{x[n+1]=x[n]-(x[n]^3+10x[n]-80)/(3x[n]^2+10)}}}

For a reasonable first guess we can use the fact that when x=3, the LHS of the original equation is 57 and when x=4, LHS=104. So let's split the difference and choose x=3.5

This gives us...


{{{x[2]=3.545454}}}
{{{x[3]=3.54499}}}
{{{x[4]=3.54499}}}

so probably close enough...