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

The solutions to the equation 2x^2+8x=-1 are x=(-4+sqrt(14))/2 ,x=-(4+sqrt(14))/2

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

The zeros of 2x^2+8x=-1 are x=(-4+sqrt(14))/2 ,x=-(4+sqrt(14))/2

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

x = \frac{- 4 + 14}{2}, x = - \frac{4 + 14}{2}

Decimal Notation

x = - 0.12917 \dots, x = - 3.87082 \dots

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 \cdot 1}{2 \cdot 2}

8^{2} - 4 \cdot 2 \cdot 1 = 214

x_{1, 2} = \frac{- 8 \pm 2 14}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 8 + 2 14}{2 \cdot 2}, x_{2} = \frac{- 8 - 2 14}{2 \cdot 2}

x = \frac{- 8 + 2 14}{2 \cdot 2} : \frac{- 4 + 14}{2}

x = \frac{- 8 - 2 14}{2 \cdot 2} : - \frac{4 + 14}{2}

The solutions to the quadratic equation are:

x = \frac{- 4 + 14}{2}, x = - \frac{4 + 14}{2}

3 x^{2} - 19 x y + 20 y^{2}

\frac{5 - 4 2}{4 - 3 2}

erf (1)

7 y^{2} - 5 y - 2

9 + x y - x - 9 y^{2}

What are the solutions to the equation 2x^2+8x=-1 ?
The solutions to the equation 2x^2+8x=-1 are x=(-4+sqrt(14))/2 ,x=-(4+sqrt(14))/2
Find the zeros of 2x^2+8x=-1
The zeros of 2x^2+8x=-1 are x=(-4+sqrt(14))/2 ,x=-(4+sqrt(14))/2