The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 0.7165\\[1 em]x_3 &= -1.1165 \end{aligned} $$Step 1:
Factor out $ \color{blue}{ x^2 }$ from $ 5x^4+2x^3-4x^2 $ and solve two separate equations:
$$ \begin{aligned} 5x^4+2x^3-4x^2 & = 0\\[1 em] \color{blue}{ x^2 }\cdot ( 5x^2+2x-4 ) & = 0 \\[1 em] \color{blue}{ x^2 = 0} ~~ \text{or} ~~ 5x^2+2x-4 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
The solutions of $ 5x^2+2x-4 = 0 $ are: $ x = -\dfrac{ 1 }{ 5 }-\dfrac{\sqrt{ 21 }}{ 5 } ~ \text{and} ~ x = -\dfrac{ 1 }{ 5 }+\dfrac{\sqrt{ 21 }}{ 5 }$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.