Find roots of polynomial $$ p(x) = 2x^2-96+288 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 4 \sqrt{ 6 }i\\[1 em]x_2 &= -4 \sqrt{ 6 }i \end{aligned} $$Explanation
Step 1:
Combine like terms:
$$ 2x^2 \color{blue}{-96} + \color{blue}{288} = 2x^2+ \color{blue}{192} $$Step 2:
The solutions of $ 2x^2+192 = 0 $ are: $ x = 4 \sqrt{ 6 } i ~ \text{and} ~ x = -4 \sqrt{ 6 } 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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