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

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

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

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

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u^{2} - 2 u - 2 = 0

u = 1 + 3, u = 1 - 3

Solution steps

u^{2} - 2 u - 2 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

u = 1 + 3, u = 1 - 3

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

(2 x - 1)^{2} = (x + 4)^{2}

x^{2} - 4.649160568 x + 3.125 = 0

v^{2} + 11 v + 30 = 0

x, f (x - 2) = x^{2} - 1

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