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

The solutions to the equation x^2-2x-2=0,x>0 are x=1+sqrt(3)

Find the zeros of x^2-2x-2=0,x>0

The zeros of x^2-2x-2=0,x>0 are x=1+sqrt(3)

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x^{2} - 2 x - 2 = 0, x > 0

x = 1 + 3

Decimal Notation

x = 2.73205 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 2) = 23

x_{1, 2} = \frac{- ( - 2 ) \pm 2 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 3}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2 3}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2 3}{2 \cdot 1} : 1 + 3

x = \frac{- ( - 2 ) - 2 3}{2 \cdot 1} : 1 - 3

The solutions to the quadratic equation are:

x = 1 + 3, x = 1 - 3

Check constrains

x = 1 + 3

y, x^{2} y^{9} = 2

x^{2} - 10 x - 9

x^{2} - 10 x - 7

(- 0.02083 x^{2}) + 4 x - 170 = 0

x^{2} - x + 2

What are the solutions to the equation x^2-2x-2=0,x>0 ?
The solutions to the equation x^2-2x-2=0,x>0 are x=1+sqrt(3)
Find the zeros of x^2-2x-2=0,x>0
The zeros of x^2-2x-2=0,x>0 are x=1+sqrt(3)