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

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

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

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

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2 \cdot x^{2} - 4 \cdot x + 1 = 0

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

Decimal Notation

x = 1.70710 \dots, x = 0.29289 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

1000 y^{9} - m^{15}

25 n^{3} + 15 n^{2} + 20 n + 12

x^{\frac{5}{2}} - 36 x^{\frac{1}{2}}

- 5 x > 20

- 2 ∣ 7 d ∣ + 9 \geq - 7

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