Question
Find roots of polynomial $$ p(x) = -\frac{49}{10}x^2-10x+100 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3.6109\\[1 em]x_2 &= -5.6518 \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 10 } $.
$$ \begin{aligned} -\frac{49}{10}x^2-10x+100 & = 0 ~~~ / \cdot \color{blue}{ 10 } \\[1 em] -49x^2-100x+1000 & = 0 \end{aligned} $$Step 2:
The solutions of $ -49x^2-100x+1000 = 0 $ are: $ x = -\dfrac{ 50 }{ 49 }-\dfrac{ 10 \sqrt{ 515}}{ 49 } ~ \text{and} ~ x = -\dfrac{ 50 }{ 49 }+\dfrac{ 10 \sqrt{ 515}}{ 49 }$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.
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