Question
Find roots of polynomial $$ p(x) = 100x^2-144x+45 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0.9815\\[1 em]x_2 &= 0.4585 \end{aligned} $$Explanation
The solutions of $ 100x^2-144x+45 = 0 $ are: $ x = \dfrac{ 18 }{ 25 }-\dfrac{ 3 \sqrt{ 19}}{ 50 } ~ \text{and} ~ x = \dfrac{ 18 }{ 25 }+\dfrac{ 3 \sqrt{ 19}}{ 50 }$.
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