Answer
$0; -\frac{12+\sqrt{15} i}{3}; \frac{-12+\sqrt{15} i}{3}$
Work Step by Step
Given the cubic equation
\begin{equation}
3 x^3+24 x^2+53 x=0.
\end{equation} First we factor out $x$: $$x(3 x^2+24 x+53 )=0.$$ We have: $$x= 0\text{ or }3 x^2+24 x+53=0.$$ 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}
a&x^2+bx+c=0\\
a & =3, b=24, c=53 \\
x & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
x & =\frac{-24 \pm \sqrt{(24)^2-4 \cdot 3\cdot(53)}}{2 \cdot 3}\\
& =\frac{-24\pm \sqrt{-60}}{6} \\
& =\frac{-24\pm 2\sqrt{15} i}{6} \\
&= \frac{-12\pm \sqrt{15} i}{3}.
\end{aligned}
\end{equation} The solution of the given equation is
\begin{equation}
\begin{aligned}
x& =0\\
x&= -\frac{12+ \sqrt{15} i}{3}\\
x& =\frac{-12+\sqrt{15} i}{3}.
\end{aligned}
\end{equation}
