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}

Decimal Notation

x = - 0.22474 \dots, x = 2.22474 \dots

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 ( - 2 ) \cdot 1}{2 ( - 2 )}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

4 - (- 4)

6 x^{2} + 19 x - 20

(3 x + y)^{2} - (2 x + 5)^{2}

x - 5 = - 2

4 m^{4} - 25

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