What are the solutions to the equation 8x(x+1)=-1 ?

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

Find the zeros of 8x(x+1)=-1

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

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8 x (x + 1) = - 1

x = \frac{- 2 + 2}{4}, x = - \frac{2 + 1}{2 2}

Solution steps

8 x (x + 1) = - 1

Expand

8 x (x + 1) : 8 x^{2} + 8 x

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

Move

1

to the left side

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

Solve with the quadratic formula

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

8^{2} - 4 \cdot 8 \cdot 1 = 42

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

Separate the solutions

x_{1} = \frac{- 8 + 4 2}{2 \cdot 8}, x_{2} = \frac{- 8 - 4 2}{2 \cdot 8}

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

x = \frac{- 8 - 4 2}{2 \cdot 8} : - \frac{2 + 1}{2 2}

The solutions to the quadratic equation are:

x = \frac{- 2 + 2}{4}, x = - \frac{2 + 1}{2 2}

- \frac{0.8 ( 10 )}{( 0.01 ( 10 ) ^{2} + 1 ) ^{2}}

4 x^{3} + 4 u

3 \leq 3 x - 4 < 2 x + 1

3 k^{2} - 21 k - 54

\frac{1}{8} + \frac{2}{3}

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