Question
Find roots of polynomial $$ p(x) = 100x^2-154x+45 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1.148\\[1 em]x_2 &= 0.392 \end{aligned} $$Explanation
The solutions of $ 100x^2-154x+45 = 0 $ are: $ x = \dfrac{ 77 }{ 100 }-\dfrac{\sqrt{ 1429 }}{ 100 } ~ \text{and} ~ x = \dfrac{ 77 }{ 100 }+\dfrac{\sqrt{ 1429 }}{ 100 }$.
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