Question
Find roots of polynomial $$ p(x) = 9x^2-18x-16 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \dfrac{ 8 }{ 3 }\\[1 em]x_2 &= -\dfrac{ 2 }{ 3 } \end{aligned} $$Explanation
The solutions of $ 9x^2-18x-16 = 0 $ are: $ x = -\dfrac{ 2 }{ 3 } ~ \text{and} ~ x = \dfrac{ 8 }{ 3 }$.
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