What are the solutions to the equation 2y^2-4y+1=0 ?

The solutions to the equation 2y^2-4y+1=0 are y=(2+sqrt(2))/2 ,y=(2-sqrt(2))/2

Find the zeros of 2y^2-4y+1=0

The zeros of 2y^2-4y+1=0 are y=(2+sqrt(2))/2 ,y=(2-sqrt(2))/2

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2 y^{2} - 4 y + 1 = 0

y = \frac{2 + 2}{2}, y = \frac{2 - 2}{2}

Decimal Notation

y = 1.70710 \dots, y = 0.29289 \dots

Solution steps

2 y^{2} - 4 y + 1 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 2 \cdot 1 = 22

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

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 2 2}{2 \cdot 2}, y_{2} = \frac{- ( - 4 ) - 2 2}{2 \cdot 2}

y = \frac{- ( - 4 ) + 2 2}{2 \cdot 2} : \frac{2 + 2}{2}

y = \frac{- ( - 4 ) - 2 2}{2 \cdot 2} : \frac{2 - 2}{2}

The solutions to the quadratic equation are:

y = \frac{2 + 2}{2}, y = \frac{2 - 2}{2}

8.1 \cdot 1 0^{- 12} = (2 x)^{2} \cdot (2.5 \cdot 1 0^{- 3})

Q, p = 6 - 6 Q + 3 Q^{2}

5 z^{2} = 17 z - 14

54000 = - 0.6 x^{2} + 360 x

(x + 1)^{2} - 7 = 0

What are the solutions to the equation 2y^2-4y+1=0 ?
The solutions to the equation 2y^2-4y+1=0 are y=(2+sqrt(2))/2 ,y=(2-sqrt(2))/2
Find the zeros of 2y^2-4y+1=0
The zeros of 2y^2-4y+1=0 are y=(2+sqrt(2))/2 ,y=(2-sqrt(2))/2