Find roots of polynomial $$ p(x) = 12x^2+x-\frac{1}{4} $$

The roots of polynomial $ p(x) $ are:

$$ \begin{aligned}x_1 &= 0.1086\\[1 em]x_2 &= -0.1919 \end{aligned} $$

Step 1:

Get rid of fractions by multipling equation by $ \color{blue}{ 4 } $.

$$ \begin{aligned} 12x^2+x-\frac{1}{4} & = 0 ~~~ / \cdot \color{blue}{ 4 } \\[1 em] 48x^2+4x-1 & = 0 \end{aligned} $$

Step 2:

The solutions of $ 48x^2+4x-1 = 0 $ are: $ x = -\dfrac{ 1 }{ 24 }-\dfrac{\sqrt{ 13 }}{ 24 } ~ \text{and} ~ x = -\dfrac{ 1 }{ 24 }+\dfrac{\sqrt{ 13 }}{ 24 }$.

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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