Answer
$x\displaystyle \in\left\{\frac{-5+\sqrt {29}}{2}, \frac{-5-\sqrt {29}}{2}, \frac{1}{2}\right\}$
Work Step by Step
See Descarte's rule of sign:
It asserts that the number of real roots is at most the number of sign changes in the sequence of polynomial's coefficients (omitting the zero coefficients), and that the difference between these two numbers is always even.
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f(x) $ with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and
$q$ is a factor of the leading coefficient, $a_{n}$.
------------------------
$f(x)=2x^{3}+9x^{2}-7x+1$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, $
$q:\qquad \pm 1, \pm 2,$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm \frac{1}{2}$
b. $f(x)=2x^{3}+9x^{2}-7x+1$, there are two sign change therefore $f(x)$ has $2$ or $0$ positive roots.
$f(-x)=-2x^{3}+9x^{2}+7x+1$, there's one sign change therefore $f(x)$ has $1$ negative root
c. Try for $x=\frac{1}{2}:$
$\begin{array}{lIII}
\underline{\frac{1}{2}}| & 2 & 9 & -7 & 1\\
& & 1 & 5 & -1\\
& -- & -- & -- & --\\
& 2 & 10 & -2 & |\underline{0}
\end{array}$
$\frac{1}{2}$ is a zero,
$f(x)=(x-\frac{1}{2})(2x^{2} +10x-2)$
d. To solve for the trinomial using the quadratic formula for the quadratic function of $ax*2+bx+c$, $x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}$...
$x=\frac{-10\pm\sqrt {10^2-4\times2\times(-2)}}{2\times2}$,
$x=\frac{-10\pm\sqrt {116}}{4}=\frac{-5\pm\sqrt {29}}{2}$
$x\displaystyle \in\left\{\frac{-5+\sqrt {29}}{2}, \frac{-5-\sqrt {29}}{2}, \frac{1}{2}\right\}$
