The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 1.4495\\[1 em]x_3 &= -3.4495 \end{aligned} $$Step 1:
Factor out $ \color{blue}{ x }$ from $ x^3+2x^2-5x $ and solve two separate equations:
$$ \begin{aligned} x^3+2x^2-5x & = 0\\[1 em] \color{blue}{ x }\cdot ( x^2+2x-5 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ x^2+2x-5 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 2:
The solutions of $ x^2+2x-5 = 0 $ are: $ x = -1-\sqrt{ 6 } ~ \text{and} ~ x = -1+\sqrt{ 6 }$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.