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

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

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

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

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

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

Decimal Notation

z = 0.25, z = - 0.33333 \dots

Solution steps

12 z^{2} + 1 \cdot z - 1 = 0

Refine

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

9 \cdot \frac{4}{3}

12 l^{2} + 48 l - 826 = 0

441

x^{2} + 11 x + 2 = 0

x^{- \frac{5}{3}}

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