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

The solutions to the equation 2x^2-4x=1 are x=(2+sqrt(6))/2 ,x=(2-sqrt(6))/2

Find the zeros of 2x^2-4x=1

The zeros of 2x^2-4x=1 are x=(2+sqrt(6))/2 ,x=(2-sqrt(6))/2

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2 x^{2} - 4 x = 1

x = \frac{2 + 6}{2}, x = \frac{2 - 6}{2}

Solution steps

2 x^{2} - 4 x = 1

Move

1

to the left side

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 2 (- 1) = 26

x_{1, 2} = \frac{- ( - 4 ) \pm 2 6}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 6}{2 \cdot 2}, x_{2} = \frac{- ( - 4 ) - 2 6}{2 \cdot 2}

x = \frac{- ( - 4 ) + 2 6}{2 \cdot 2} : \frac{2 + 6}{2}

x = \frac{- ( - 4 ) - 2 6}{2 \cdot 2} : \frac{2 - 6}{2}

The solutions to the quadratic equation are:

x = \frac{2 + 6}{2}, x = \frac{2 - 6}{2}

1 + 7 x = 9 + 5 x

20 n^{2} - 200 n

4 k^{3} + 12 k^{2} + 11 k + 3

\frac{1}{∣ x - 1 ∣} < \frac{1}{∣ x + 2 ∣}

220 - 20 + 320 - 245

What are the solutions to the equation 2x^2-4x=1 ?
The solutions to the equation 2x^2-4x=1 are x=(2+sqrt(6))/2 ,x=(2-sqrt(6))/2
Find the zeros of 2x^2-4x=1
The zeros of 2x^2-4x=1 are x=(2+sqrt(6))/2 ,x=(2-sqrt(6))/2