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

The solutions to the equation 0=-2x^2+4x+1 are x=-(-2+sqrt(6))/2 ,x=(2+sqrt(6))/2

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

The zeros of 0=-2x^2+4x+1 are x=-(-2+sqrt(6))/2 ,x=(2+sqrt(6))/2

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0 = - 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

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

Switch sides

- 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}

- 3 x^{2} + 16 x - 10 = 0

(x - \frac{3}{2})^{2} = \frac{25}{4}

x^{2} - 25 x + 75 = 0

x^{2} - 8 x + 13

x^{2} - 8 x + 11

What are the solutions to the equation 0=-2x^2+4x+1 ?
The solutions to the equation 0=-2x^2+4x+1 are x=-(-2+sqrt(6))/2 ,x=(2+sqrt(6))/2
Find the zeros of 0=-2x^2+4x+1
The zeros of 0=-2x^2+4x+1 are x=-(-2+sqrt(6))/2 ,x=(2+sqrt(6))/2