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=(5-sqrt(41))/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=(5-sqrt(41))/8

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

x = \frac{5 + 41}{8}, x = \frac{5 - 41}{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 \cdot 4 ( - 1 )}{2 \cdot 4}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

∣ x + 2 ∣ \geq 4

4 x^{2} + 12 x - 20 = 0

- 1 + 6

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=(5-sqrt(41))/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=(5-sqrt(41))/8