Question
Find roots of polynomial $$ p(x) = -\frac{49}{10}x^2-20x+31 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1.1982\\[1 em]x_2 &= -5.2799 \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 10 } $.
$$ \begin{aligned} -\frac{49}{10}x^2-20x+31 & = 0 ~~~ / \cdot \color{blue}{ 10 } \\[1 em] -49x^2-200x+310 & = 0 \end{aligned} $$Step 2:
The solutions of $ -49x^2-200x+310 = 0 $ are: $ x = -\dfrac{ 100 }{ 49 }-\dfrac{\sqrt{ 25190 }}{ 49 } ~ \text{and} ~ x = -\dfrac{ 100 }{ 49 }+\dfrac{\sqrt{ 25190 }}{ 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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