Question
Find roots of polynomial $$ p(x) = 16+x-2x^2 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3.0895\\[1 em]x_2 &= -2.5895 \end{aligned} $$Explanation
Step 1:
Write polynomial in descending order
$$ \begin{aligned} 16+x-2x^2 & = 0\\[1 em] -2x^2+x+16 & = 0 \end{aligned} $$Step 2:
The solutions of $ -2x^2+x+16 = 0 $ are: $ x = \dfrac{ 1 }{ 4 }-\dfrac{\sqrt{ 129 }}{ 4 } ~ \text{and} ~ x = \dfrac{ 1 }{ 4 }+\dfrac{\sqrt{ 129 }}{ 4 }$.
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