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=(sqrt(5)-2)/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=(sqrt(5)-2)/2

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

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

Decimal Notation

x = - 2.11803 \dots, x = 0.11803 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

x = \frac{- ( - 8 ) - 4 5}{2 ( - 4 )} : \frac{5 - 2}{2}

The solutions to the quadratic equation are:

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

x, x^{2} - 6 x - 3 = 0

x (x - 4) = 10

12 x^{2} - 60 x + 50 = 0

2 x^{2} - 5 x + 2

3 x^{2} + 6 x - 60

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=(sqrt(5)-2)/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=(sqrt(5)-2)/2