Question
Find roots of polynomial $$ p(x) = y^2+y+\frac{13}{4} $$
Answer
The roots of polynomial $ p(y) $ are:
$$ \begin{aligned}y_1 &= -\dfrac{ 1 }{ 2 }+\sqrt{ 3 }i\\[1 em]y_2 &= -\dfrac{ 1 }{ 2 }-\sqrt{ 3 }i \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 4 } $.
$$ \begin{aligned} y^2+y+\frac{13}{4} & = 0 ~~~ / \cdot \color{blue}{ 4 } \\[1 em] 4y^2+4y+13 & = 0 \end{aligned} $$Step 2:
The solutions of $ 4y^2+4y+13 = 0 $ are: $ y = -\dfrac{ 1 }{ 2 }+\sqrt{ 3 }i ~ \text{and} ~ y = -\dfrac{ 1 }{ 2 }-\sqrt{ 3 }i$.
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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