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

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

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

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

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

x = 1 + 5, x = 1 - 5

Decimal Notation

x = 3.23606 \dots, x = - 1.23606 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 4) = 25

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + 5, x = 1 - 5

(\frac{4}{9})^{\frac{3}{2}}

25 \cdot 55

4 x + 3 x - 7 = 60 - 2 x - 13

7 (y + 4.5) = 32.5

x - 4 = 4

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