What are the solutions to the equation -4x^2-5x+1=0 ?

The solutions to the equation -4x^2-5x+1=0 are x=-(5+sqrt(41))/8 ,x=(sqrt(41)-5)/8

Find the zeros of -4x^2-5x+1=0

The zeros of -4x^2-5x+1=0 are x=-(5+sqrt(41))/8 ,x=(sqrt(41)-5)/8

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- 4 x^{2} - 5 x + 1 = 0

x = - \frac{5 + 41}{8}, x = \frac{41 - 5}{8}

Decimal Notation

x = - 1.42539 \dots, x = 0.17539 \dots

Solution steps

- 4 x^{2} - 5 x + 1 = 0

Solve with the quadratic formula

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

(- 5)^{2} - 4 (- 4) \cdot 1 = 41

x_{1, 2} = \frac{- ( - 5 ) \pm 41}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 41}{2 ( - 4 )}, x_{2} = \frac{- ( - 5 ) - 41}{2 ( - 4 )}

x = \frac{- ( - 5 ) + 41}{2 ( - 4 )} : - \frac{5 + 41}{8}

x = \frac{- ( - 5 ) - 41}{2 ( - 4 )} : \frac{41 - 5}{8}

The solutions to the quadratic equation are:

x = - \frac{5 + 41}{8}, x = \frac{41 - 5}{8}

7 x^{2} - (2 x - 3)^{2} = 3 (4 x + 1)

12 x^{2} - 80 x + 100 = 0

x^{2} + 12 x - 13 = 0

- 18 p = p^{2} + 81

r^{2} - r + 1 = 0

What are the solutions to the equation -4x^2-5x+1=0 ?
The solutions to the equation -4x^2-5x+1=0 are x=-(5+sqrt(41))/8 ,x=(sqrt(41)-5)/8
Find the zeros of -4x^2-5x+1=0
The zeros of -4x^2-5x+1=0 are x=-(5+sqrt(41))/8 ,x=(sqrt(41)-5)/8