What are the solutions to the equation 8x^2+8x+1=0 ?

The solutions to the equation 8x^2+8x+1=0 are x=(-2+sqrt(2))/4 ,x=-(sqrt(2)+1)/(2sqrt(2))

Find the zeros of 8x^2+8x+1=0

The zeros of 8x^2+8x+1=0 are x=(-2+sqrt(2))/4 ,x=-(sqrt(2)+1)/(2sqrt(2))

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

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

Decimal Notation

x = - 0.14644 \dots, x = - 0.85355 \dots

Solution steps

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{2 b + 7}{9 b} + \frac{b + 4}{6 b}

0 = 3 x^{2} - 8 x + 5

lo g_{6} (x^{2} - 5 x) > 2

7 t^{2} + 28 t + 56 = 0

3.75 - 3 x \leq 6

What are the solutions to the equation 8x^2+8x+1=0 ?
The solutions to the equation 8x^2+8x+1=0 are x=(-2+sqrt(2))/4 ,x=-(sqrt(2)+1)/(2sqrt(2))
Find the zeros of 8x^2+8x+1=0
The zeros of 8x^2+8x+1=0 are x=(-2+sqrt(2))/4 ,x=-(sqrt(2)+1)/(2sqrt(2))