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 \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}

\frac{1}{2} + 6

(27 a^{6} b^{- 3})^{\frac{2}{3}}

2 x + 3 = 3 + 2 x

\frac{- 1}{2} \cdot 2

\frac{c + \frac{125}{c ^{2}}}{1 + \frac{5}{c}}

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