What are the solutions to the equation x^2+x/(12)-1/12 =0 ?

The solutions to the equation x^2+x/(12)-1/12 =0 are x= 1/4 ,x=-1/3

Find the zeros of x^2+x/(12)-1/12 =0

The zeros of x^2+x/(12)-1/12 =0 are x= 1/4 ,x=-1/3

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x^{2} + \frac{x}{12} - \frac{1}{12} = 0

x = \frac{1}{4}, x = - \frac{1}{3}

Decimal Notation

x = 0.25, x = - 0.33333 \dots

Solution steps

x^{2} + \frac{x}{12} - \frac{1}{12} = 0

Multiply both sides by

12

12 x^{2} + x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 12 ( - 1 )}{2 \cdot 12}

1^{2} - 4 \cdot 12 (- 1) = 7

x_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 1 + 7}{2 \cdot 12}, x_{2} = \frac{- 1 - 7}{2 \cdot 12}

x = \frac{- 1 + 7}{2 \cdot 12} : \frac{1}{4}

x = \frac{- 1 - 7}{2 \cdot 12} : - \frac{1}{3}

The solutions to the quadratic equation are:

x = \frac{1}{4}, x = - \frac{1}{3}

k + 3 \leq - 1

3 x^{2} + 12 x - 288 = 0

\frac{f}{5} + 2 = 8

(- 3 m^{5}) (- 2 m^{4})

6 + \frac{2 x + 4}{3} x = 8

What are the solutions to the equation x^2+x/(12)-1/12 =0 ?
The solutions to the equation x^2+x/(12)-1/12 =0 are x= 1/4 ,x=-1/3
Find the zeros of x^2+x/(12)-1/12 =0
The zeros of x^2+x/(12)-1/12 =0 are x= 1/4 ,x=-1/3