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

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

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

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

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

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

Solution steps

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

Switch sides

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}

- 2 x - 6 = 18

(- 27) \frac{4}{3}

x^{2} + 14 x - 24 = 0

j + 17 = 20

2 (\frac{9}{25})

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