$$r(3r^2+7r-8) = 0$$
Answer
$$ \begin{matrix}r_1 = 0 & r_2 = -\dfrac{ 7 }{ 6 }-\dfrac{\sqrt{ 145 }}{ 6 } & r_3 = -\dfrac{ 7 }{ 6 }+\dfrac{\sqrt{ 145 }}{ 6 } \end{matrix} $$
Explanation
$$ \begin{aligned} r(3r^2+7r-8) &= 0&& \text{simplify left side} \\[1 em]3r^3+7r^2-8r &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 3x^{3}+7x^{2}-8x = 0 } $, first we need to factor our $ x $.
$$ 3x^{3}+7x^{2}-8x = x \left( 3x^{2}+7x-8 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 3x^{2}+7x-8 = 0$.
$ 3x^{2}+7x-8 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
This page was created using
Equations Solver