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

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

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

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

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

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

- (x - (1 0^{- 5}))^{2} = (3.12 \cdot 1 0^{- 5})

6 x (x - 1) = 2 - 5 x

3 x - 10 < 5 x - 2

820

5 ∣ - 4 x + 2 ∣ - 3 \leq 27

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