Question
Find roots of polynomial $$ p(x) = 24x^2+191x-504 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 2.0899\\[1 em]x_2 &= -10.0482 \end{aligned} $$Explanation
The solutions of $ 24x^2+191x-504 = 0 $ are: $ x = -\dfrac{ 191 }{ 48 }-\dfrac{\sqrt{ 84865 }}{ 48 } ~ \text{and} ~ x = -\dfrac{ 191 }{ 48 }+\dfrac{\sqrt{ 84865 }}{ 48 }$.
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