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

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

Find the zeros of 12y^2+2y-2=0

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

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12 y^{2} + 2 y - 2 = 0

y = \frac{1}{3}, y = - \frac{1}{2}

Decimal Notation

y = 0.33333 \dots, y = - 0.5

Solution steps

12 y^{2} + 2 y - 2 = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 12 (- 2) = 10

y_{1, 2} = \frac{- 2 \pm 10}{2 \cdot 12}

Separate the solutions

y_{1} = \frac{- 2 + 10}{2 \cdot 12}, y_{2} = \frac{- 2 - 10}{2 \cdot 12}

y = \frac{- 2 + 10}{2 \cdot 12} : \frac{1}{3}

y = \frac{- 2 - 10}{2 \cdot 12} : - \frac{1}{2}

The solutions to the quadratic equation are:

y = \frac{1}{3}, y = - \frac{1}{2}

q^{2} + 4 q + 4 = 0

(2 x + 3)^{2} - 4 (x - 1) (x + 3) = x^{2}

x^{2} - 2 x - 80 = 0

a - 5 - a^{2} = 0

x^{2} - x = - 0.25

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