Find roots of polynomial $$ p(x) = 20x^3+9x^2-2x $$

The roots of polynomial $ p(x) $ are:

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 0.1631\\[1 em]x_3 &= -0.6131 \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ x }$ from $ 20x^3+9x^2-2x $ and solve two separate equations:

$$ \begin{aligned} 20x^3+9x^2-2x & = 0\\[1 em] \color{blue}{ x }\cdot ( 20x^2+9x-2 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ 20x^2+9x-2 & = 0 \end{aligned} $$

One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.

Step 2:

The solutions of $ 20x^2+9x-2 = 0 $ are: $ x = -\dfrac{ 9 }{ 40 }-\dfrac{\sqrt{ 241 }}{ 40 } ~ \text{and} ~ x = -\dfrac{ 9 }{ 40 }+\dfrac{\sqrt{ 241 }}{ 40 }$.

You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.

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