Find roots of polynomial $$ p(x) = 150x^2+130x+1056 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -\frac{ 13 }{ 30 }+\frac{\sqrt{ 6167 }}{ 30 }i\\[1 em]x_2 &= -\frac{ 13 }{ 30 }- \frac{\sqrt{ 6167 }}{ 30 }i \end{aligned} $$Explanation
The solutions of $ 150x^2+130x+1056 = 0 $ are: $ x = -\dfrac{ 13 }{ 30 }+\dfrac{\sqrt{ 6167 }}{ 30 }i ~ \text{and} ~ x = -\dfrac{ 13 }{ 30 }-\dfrac{\sqrt{ 6167 }}{ 30 }i$.
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