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

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

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

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

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x^{2} - 2 x = 2

x = 1 + 3, x = 1 - 3

Solution steps

x^{2} - 2 x = 2

Move

2

to the left side

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

\frac{3}{3 x y} + \frac{2 y}{6 x}

9 x^{2} - 3 x - 4 = 0

90 = 66 e^{- 0.0174 (90)} + 75

x^{2} + 11 x + 28 < 0

50 - 98 x^{2}

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