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

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

Find the zeros of r^2-2r-2=0

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

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r^{2} - 2 r - 2 = 0

r = 1 + 3, r = 1 - 3

Decimal Notation

r = 2.73205 \dots, r = - 0.73205 \dots

Solution steps

r^{2} - 2 r - 2 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

r = 1 + 3, r = 1 - 3

x^{2} + 6 x = 5

y^{2} + 12 y + 20

x^{2} + 2 x - 3

2 + 3 ∣ x - 9 ∣ = 8

20 x^{4} + 45 x^{5}

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