What are the solutions to the equation x(2x-2)=2 ?

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

Find the zeros of x(2x-2)=2

The zeros of x(2x-2)=2 are x=(1+sqrt(5))/2 ,x=(1-sqrt(5))/2

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x (2 x - 2) = 2

x = \frac{1 + 5}{2}, x = \frac{1 - 5}{2}

Decimal Notation

x = 1.61803 \dots, x = - 0.61803 \dots

Solution steps

x (2 x - 2) = 2

Expand

x (2 x - 2) : 2 x^{2} - 2 x

2 x^{2} - 2 x = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{1 + 5}{2}, x = \frac{1 - 5}{2}

x^{2} + 14 x = - 13

3 x^{2} - 3 x + \frac{3}{4} = 0

\frac{1}{3} x^{2} + \frac{1}{4} x - 3 = 0

x, x^{2} + x y + y^{2} = 25

x, x^{2} + x y + y^{2} = 36

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