What are the solutions to the equation 12x^2+7x+1=0 ?

The solutions to the equation 12x^2+7x+1=0 are x=-1/4 ,x=-1/3

Find the zeros of 12x^2+7x+1=0

The zeros of 12x^2+7x+1=0 are x=-1/4 ,x=-1/3

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12 x^{2} + 7 x + 1 = 0

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

Solution steps

12 x^{2} + 7 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 12 \cdot 1}{2 \cdot 12}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

x^{2} - 18 x + 68 = 0

5 x - 7 = - 22

2 - \frac{4}{9}

7 x - x^{2} = 10

49 x^{2} - 64

What are the solutions to the equation 12x^2+7x+1=0 ?
The solutions to the equation 12x^2+7x+1=0 are x=-1/4 ,x=-1/3
Find the zeros of 12x^2+7x+1=0
The zeros of 12x^2+7x+1=0 are x=-1/4 ,x=-1/3