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

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

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

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

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

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

Decimal Notation

d = 1.70710 \dots, d = 0.29289 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

2 ∣ 3 x - 5 ∣ \geq 4

∣ x ∣ = 2

- 8 m^{2} + 8 m - 2

2 \cdot (\frac{1}{2})^{1}

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