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=-(sqrt(2)+1)/(sqrt(2))

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

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

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

x = \frac{- 2 + 2}{2}, x = - \frac{2 + 1}{2}

Decimal Notation

x = - 0.29289 \dots, x = - 1.70710 \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 \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 + 1}{2}

The solutions to the quadratic equation are:

x = \frac{- 2 + 2}{2}, x = - \frac{2 + 1}{2}

4 x^{2} + 16 x + 19 = 0

- x^{2} - 3 x = 0

20 x^{2} - 5 = 21 x

(x - 1) (x + 2) - (2 x - 3) (x + 4) - x + 14 = 0

(2 x + 3)^{2} + 5 = 8

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