The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 0.6866\\[1 em]x_3 &= -0.6866\\[1 em]x_4 &= 0.6866i\\[1 em]x_5 &= -0.6866i \end{aligned} $$Step 1:
Factor out $ \color{blue}{ 2x }$ from $ 18x^5-4x $ and solve two separate equations:
$$ \begin{aligned} 18x^5-4x & = 0\\[1 em] \color{blue}{ 2x }\cdot ( 9x^4-2 ) & = 0 \\[1 em] \color{blue}{ 2x = 0} ~~ \text{or} ~~ 9x^4-2 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
Polynomial $ 9x^4-2 $ has no rational roots that can be found using Rational Root Test, so the roots were found using quartic formulas.