Answer
$x=0$
$x=1$
$x= -\frac{7}{5}$
Work Step by Step
Given the cubic equation
\begin{equation}
5 x^3+2 x^2-7 x=0.
\end{equation} First we factor out $x$. $$x(5x^2+2x-7)=0.$$ One of the solutions is $$x=0.$$ The quadratic quation can be best solved by the quadratic formula.
\begin{equation}
\begin{aligned}
5x^2+2 x-7 &=0\\
a & =5, b=2, c=-7 \\
x & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
x & =\frac{-2 \pm \sqrt{2^2-4 \cdot 5\cdot(-7)}}{2 \cdot 5}\\
& =\frac{-2\pm \sqrt{144}}{10} \\
& =\frac{-2\pm 12}{10} \\
& =\frac{-1\pm 6}{5}.
\end{aligned}
\end{equation} The solution of the given equation is \begin{equation}
\begin{aligned}
x&= 0\\
x& =\frac{-1+ 6}{5}= 1\\
x& =\frac{-1- 6}{5}= -\frac{7}{5}.
\end{aligned}
\end{equation}
