Find roots of polynomial $$ p(x) = 18x^2+26x-40 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0.9342\\[1 em]x_2 &= -2.3787 \end{aligned} $$Explanation
The solutions of $ 18x^2+26x-40 = 0 $ are: $ x = -\dfrac{ 13 }{ 18 }-\dfrac{\sqrt{ 889 }}{ 18 } ~ \text{and} ~ x = -\dfrac{ 13 }{ 18 }+\dfrac{\sqrt{ 889 }}{ 18 }$.
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