What are the solutions to the equation 2x^2+8x=1 ?

The solutions to the equation 2x^2+8x=1 are x=(-4+3sqrt(2))/2 ,x=-(4+3sqrt(2))/2

Find the zeros of 2x^2+8x=1

The zeros of 2x^2+8x=1 are x=(-4+3sqrt(2))/2 ,x=-(4+3sqrt(2))/2

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2 x^{2} + 8 x = 1

x = \frac{- 4 + 3 2}{2}, x = - \frac{4 + 3 2}{2}

Solution steps

2 x^{2} + 8 x = 1

Move

1

to the left side

2 x^{2} + 8 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

8^{2} - 4 \cdot 2 (- 1) = 62

x_{1, 2} = \frac{- 8 \pm 6 2}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 8 + 6 2}{2 \cdot 2}, x_{2} = \frac{- 8 - 6 2}{2 \cdot 2}

x = \frac{- 8 + 6 2}{2 \cdot 2} : \frac{- 4 + 3 2}{2}

x = \frac{- 8 - 6 2}{2 \cdot 2} : - \frac{4 + 3 2}{2}

The solutions to the quadratic equation are:

x = \frac{- 4 + 3 2}{2}, x = - \frac{4 + 3 2}{2}

x^{- 8}

2 x^{2} - 13 x y - 24 y^{2}

1 - x + 2 a (1 - x)

2 x^{2} - 9 x - 35

9 (2 x - 3)^{2} + 7 = 448

What are the solutions to the equation 2x^2+8x=1 ?
The solutions to the equation 2x^2+8x=1 are x=(-4+3sqrt(2))/2 ,x=-(4+3sqrt(2))/2
Find the zeros of 2x^2+8x=1
The zeros of 2x^2+8x=1 are x=(-4+3sqrt(2))/2 ,x=-(4+3sqrt(2))/2