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

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

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

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

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

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

Decimal Notation

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

Solution steps

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

Switch sides

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}

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

∣ 2 x + 3 ∣ = 10

(\frac{5}{8})^{3}

35 b = 7

- 7 x - 3 < - 45

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