Find roots of polynomial $$ p(x) = 4r^2-64r+255 $$
Answer
The roots of polynomial $ p(r) $ are:
$$ \begin{aligned}r_1 &= \frac{ 17 }{ 2 }\\[1 em]r_2 &= \frac{ 15 }{ 2 } \end{aligned} $$Explanation
The solutions of $ 4r^2-64r+255 = 0 $ are: $ r = \dfrac{ 15 }{ 2 } ~ \text{and} ~ r = \dfrac{ 17 }{ 2 }$.
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