Question 821307
{{{x^2-3x^5>10}}}
{{{-3x^5+x^2-10>0}}}
{{{3x^5-x^2+10<0}}}
A graphing calculator would tell you that
{{{f(x)=3x^5-x^2+10=0}}} happens for {{{x=-1.2311}}} (rounded),
with the {{{3x^5-x^2+10<0}}} only for {{{x<-1.2311}}} .
Before graphing calculators, I would have calculated the derivative as
{{{df/dx=15x^4-2x=15x(x^3-2/15)}}}={{{15x(x-root(3,2/15))(x^2+root(3,2/15)x+(root(3,2/15))^2)}}}
Since the last factor is always positive, the zeros of the derivative are at {{{x=0}}} and at {{{x=root(3,2/15)}}} . 
The derivative is negative in between those two values of {{{x}}} ,
meaning that is an interval where {{{f(x)}}} decreases.
For other values of {{{x}}} , the derivative is positive and the function increases.
At {{{x=0}}} where {{{f(0)=10}}} we have a maximum of {{{f(x)}}} ,
and at {{{x=root(3,2/15)}}} , a minimum of {{{f(x)}}} , where {{{f(x)>0}}} .
Since {{{f(-2)=-90}}} and {{{f(-1)=6}}} ,
The value of {{{x}}} that makes {{{f(x)=0}}} is in between, {{{-2<x<-1}}} .
At that point we would try guess-and-check values in between, aiming to get closer limits on {{{x}}}, maybe going through
{{{-1.5<x<-1}}} , {{{-1.5<x<-1.2}}} , {{{-1.3<x<-1}}} and so on, until I got a close enough approximation.