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

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

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

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

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

x = \frac{5 + 41}{2}, x = \frac{5 - 41}{2}

Decimal Notation

x = 5.70156 \dots, x = - 0.70156 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

x = \frac{- ( - 5 ) + 41}{2 \cdot 1} : \frac{5 + 41}{2}

x = \frac{- ( - 5 ) - 41}{2 \cdot 1} : \frac{5 - 41}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 41}{2}, x = \frac{5 - 41}{2}

z, - 7 z^{2} = - 28

x, 45 = x \cdot 180

x, ∣ 2 x - 3 ∣ < 3

\frac{25}{π}

x, - 4 (3 x + 1) + x - 3 = 15

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