Question
Find roots of polynomial $$ p(x) = -\frac{49}{10}t^2+\frac{688}{100}t-\frac{65}{100} $$
Answer
The roots of polynomial $ p(t) $ are:
$$ \begin{aligned}t_1 &= 1.3022\\[1 em]t_2 &= 0.1019 \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 100 } $.
$$ \begin{aligned} -\frac{49}{10}t^2+\frac{688}{100}t-\frac{65}{100} & = 0 ~~~ / \cdot \color{blue}{ 100 } \\[1 em] -490t^2+688t-65 & = 0 \end{aligned} $$Step 2:
The solutions of $ -490t^2+688t-65 = 0 $ are: $ t = \dfrac{ 172 }{ 245 }-\dfrac{\sqrt{ 86486 }}{ 490 } ~ \text{and} ~ t = \dfrac{ 172 }{ 245 }+\dfrac{\sqrt{ 86486 }}{ 490 }$.
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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