What are the solutions to the equation 4x^2+8x-1=0 ?

The solutions to the equation 4x^2+8x-1=0 are x=(-2+sqrt(5))/2 ,x=-(2+sqrt(5))/2

Find the zeros of 4x^2+8x-1=0

The zeros of 4x^2+8x-1=0 are x=(-2+sqrt(5))/2 ,x=-(2+sqrt(5))/2

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

x = \frac{- 2 + 5}{2}, x = - \frac{2 + 5}{2}

Solution steps

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

Solve with the quadratic formula

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

8^{2} - 4 \cdot 4 (- 1) = 45

x_{1, 2} = \frac{- 8 \pm 4 5}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 8 + 4 5}{2 \cdot 4}, x_{2} = \frac{- 8 - 4 5}{2 \cdot 4}

x = \frac{- 8 + 4 5}{2 \cdot 4} : \frac{- 2 + 5}{2}

x = \frac{- 8 - 4 5}{2 \cdot 4} : - \frac{2 + 5}{2}

The solutions to the quadratic equation are:

x = \frac{- 2 + 5}{2}, x = - \frac{2 + 5}{2}

2 x^{2} + 5 x + 3 \leq 0

n^{2} - 12 n + 27 = 0

\frac{2}{3} s + 12 = \frac{- 2}{3} s + 14

- \frac{7}{8} - \frac{1}{2}

\frac{7 x - 1}{6} - \frac{2 x - 1}{2} \leq \frac{1}{2}

What are the solutions to the equation 4x^2+8x-1=0 ?
The solutions to the equation 4x^2+8x-1=0 are x=(-2+sqrt(5))/2 ,x=-(2+sqrt(5))/2
Find the zeros of 4x^2+8x-1=0
The zeros of 4x^2+8x-1=0 are x=(-2+sqrt(5))/2 ,x=-(2+sqrt(5))/2