Find roots of polynomial $$ p(x) = 18x^2-9x-1 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0.5936\\[1 em]x_2 &= -0.0936 \end{aligned} $$Explanation
The solutions of $ 18x^2-9x-1 = 0 $ are: $ x = \dfrac{ 1 }{ 4 }-\dfrac{\sqrt{ 17 }}{ 12 } ~ \text{and} ~ x = \dfrac{ 1 }{ 4 }+\dfrac{\sqrt{ 17 }}{ 12 }$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.
This page was created using
Polynomial Roots Calculator