Find roots of polynomial $$ p(x) = 16x^2-9x+1 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0.4101\\[1 em]x_2 &= 0.1524 \end{aligned} $$Explanation
The solutions of $ 16x^2-9x+1 = 0 $ are: $ x = \dfrac{ 9 }{ 32 }-\dfrac{\sqrt{ 17 }}{ 32 } ~ \text{and} ~ x = \dfrac{ 9 }{ 32 }+\dfrac{\sqrt{ 17 }}{ 32 }$.
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