Question
Find roots of polynomial $$ p(x) = -45x^2-92x+120 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0.9043\\[1 em]x_2 &= -2.9488 \end{aligned} $$Explanation
The solutions of $ -45x^2-92x+120 = 0 $ are: $ x = -\dfrac{ 46 }{ 45 }-\dfrac{ 2 \sqrt{ 1879}}{ 45 } ~ \text{and} ~ x = -\dfrac{ 46 }{ 45 }+\dfrac{ 2 \sqrt{ 1879}}{ 45 }$.
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