Answer
$-\frac{1}{2}- \sqrt{2}; 0; -\frac{1}{2}+ \sqrt{2}$
Work Step by Step
Given the cubic equation \begin{equation}
8 x^3+4 x^2-14 x=0.
\end{equation} First we factor out $2x$: $$2x(4x^2+2x-7)=0.$$ We have $$x=0\text{ or } 4x^2+2x-7=0.$$ The quadratic equation can be best solved by the quadratic formula. We identify the constants $a$, $b$, $c$ from the general form of the equation and solve it using the quadratic formula:
\begin{equation}
\begin{aligned}
ax^2+bx+c&=0\\
a & =4, b=2, c=-7 \\
x & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
x & =\frac{-4 \pm \sqrt{4^2-4 \cdot 4\cdot(-7)}}{2 \cdot 4}\\
& =\frac{-4\pm \sqrt{128}}{8} \\
& =\frac{-4\pm \sqrt{64\cdot 2}}{8} \\
& =\frac{-4\pm 8 \sqrt{ 2}}{8} \\
&= -\frac{1}{2}\pm \sqrt{2}.
\end{aligned}
\end{equation} The solution of the given equation is
\begin{equation}
\begin{aligned}
x& =-\frac{1}{2}- \sqrt{2}\\
x&= 0\\
x& =-\frac{1}{2}+ \sqrt{2}.
\end{aligned}
\end{equation}
