The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= \frac{ 13 }{ 6 }\\[1 em]x_3 &= \frac{ 1 }{ 3 } \end{aligned} $$Step 1:
Factor out $ \color{blue}{ 2x }$ from $ 36x^3-90x^2+26x $ and solve two separate equations:
$$ \begin{aligned} 36x^3-90x^2+26x & = 0\\[1 em] \color{blue}{ 2x }\cdot ( 18x^2-45x+13 ) & = 0 \\[1 em] \color{blue}{ 2x = 0} ~~ \text{or} ~~ 18x^2-45x+13 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
The solutions of $ 18x^2-45x+13 = 0 $ are: $ x = \dfrac{ 1 }{ 3 } ~ \text{and} ~ x = \dfrac{ 13 }{ 6 }$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.