Question
Find roots of polynomial $$ p(x) = \cancel{18}-x -\cancel{\frac{18}{100}}-x^2 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3.7509\\[1 em]x_2 &= -4.7509 \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 100 } $.
$$ \begin{aligned} \cancel{18}-x -\cancel{\frac{18}{100}}-x^2 & = 0 ~~~ / \cdot \color{blue}{ 100 } \\[1 em] 1800-100x-18-100x^2 & = 0 \end{aligned} $$Step 2:
Combine like terms:
$$ \color{blue}{1800} -100x \color{blue}{-18} -100x^2 = -100x^2-100x+ \color{blue}{1782} $$Step 3:
The solutions of $ -100x^2-100x+1782 = 0 $ are: $ x = -\dfrac{ 1 }{ 2 }-\dfrac{\sqrt{ 1807 }}{ 10 } ~ \text{and} ~ x = -\dfrac{ 1 }{ 2 }+\dfrac{\sqrt{ 1807 }}{ 10 }$.
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