What are the solutions to the equation 2x^2-2x=1 ?

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

Find the zeros of 2x^2-2x=1

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

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2 x^{2} - 2 x = 1

x = \frac{1 + 3}{2}, x = \frac{1 - 3}{2}

Decimal Notation

x = 1.36602 \dots, x = - 0.36602 \dots

Solution steps

2 x^{2} - 2 x = 1

Move

1

to the left side

2 x^{2} - 2 x - 1 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{1 + 3}{2}, x = \frac{1 - 3}{2}

- \frac{1}{2} \leq \frac{4 \cdot 3 x}{5} \leq \frac{1}{4}

8 (\frac{1}{4}) - 1 + 0.5 (10)

\frac{10 !}{3 ! ( 10 - 3 ) !}

100 n^{2} - 10 = 15

4 (3 + x) = 36

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