#### Answer

Solution set: $(-\displaystyle \infty, -\frac{1}{2})\cup(\frac{1}{3},\infty)$

#### Work Step by Step

Follow the "Procedure for Solving Polynomial lnequalities",\ p.412:
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a polynomial function.
$6x^{2}+x >1$
$6x^{2}+x-1>0$
$f(x)= 6x^{2}+x-1$
factor the trinomial...
find factors of $6(-1)=-6$ that add to $+1:$
($3$ and $-2$ )
$6x^{2}+x-1=6x^{2}+3x-2x-1= \quad$ ... in pairs ...
$=3x(2x+1)-(2x+1)=(3x-1)(2x+1)$
$f(x)=(3x-1)(2x+1)>0$
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$(3x-1)(2x+1)=0$
$x=-\displaystyle \frac{1}{2} x=\displaystyle \frac{1}{3}$
3. Locate these boundary points on a number line, thereby dividing the number line into intervals.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & (-\infty, -\frac{1}{2}) & (-\frac{1}{2},\frac{1}{3}) & (\frac{1}{3},\infty)\\
a=test.val. & -1 & 0 & 1\\
f(a) & (-4)(-1) & (-1)(1) & (2)(3)\\
f(a) > 0 ? & T & F & T
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $(-\displaystyle \infty, -\frac{1}{2})\cup(\frac{1}{3},\infty)$