What are the solutions to the equation 11y^2-12y+2=0 ?

The solutions to the equation 11y^2-12y+2=0 are y=(6+sqrt(14))/(11),y=(6-sqrt(14))/(11)

Find the zeros of 11y^2-12y+2=0

The zeros of 11y^2-12y+2=0 are y=(6+sqrt(14))/(11),y=(6-sqrt(14))/(11)

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

y = \frac{6 + 14}{11}, y = \frac{6 - 14}{11}

Decimal Notation

y = 0.88560 \dots, y = 0.20530 \dots

Solution steps

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

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 11 \cdot 2}{2 \cdot 11}

(- 12)^{2} - 4 \cdot 11 \cdot 2 = 214

y_{1, 2} = \frac{- ( - 12 ) \pm 2 14}{2 \cdot 11}

Separate the solutions

y_{1} = \frac{- ( - 12 ) + 2 14}{2 \cdot 11}, y_{2} = \frac{- ( - 12 ) - 2 14}{2 \cdot 11}

y = \frac{- ( - 12 ) + 2 14}{2 \cdot 11} : \frac{6 + 14}{11}

y = \frac{- ( - 12 ) - 2 14}{2 \cdot 11} : \frac{6 - 14}{11}

The solutions to the quadratic equation are:

y = \frac{6 + 14}{11}, y = \frac{6 - 14}{11}

- 2 x^{2} - 2 x + 40 = 0

x^{2} = 69

x, x^{2} y^{9} = 2

3 x^{2} + 6 x - 4 = 2 x + 3

x (2 + 2 x) = 60

What are the solutions to the equation 11y^2-12y+2=0 ?
The solutions to the equation 11y^2-12y+2=0 are y=(6+sqrt(14))/(11),y=(6-sqrt(14))/(11)
Find the zeros of 11y^2-12y+2=0
The zeros of 11y^2-12y+2=0 are y=(6+sqrt(14))/(11),y=(6-sqrt(14))/(11)