What are the solutions to the equation solvefor x,f=2x^2-2x+1 ?

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

Find the zeros of solvefor x,f=2x^2-2x+1

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

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solve for

x, f = 2 x^{2} - 2 x + 1

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

Solution steps

f = 2 x^{2} - 2 x + 1

Switch sides

2 x^{2} - 2 x + 1 = f

Move

f

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 2 (1 - f) : 2 2 f - 1

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

2 (x - 2)^{2} - (3 x - 6) = 9

x^{2} + 6 x - 100 = 0

x^{2} + 1.6 x = 0.36

(x - 4) \cdot (x + 2) = 0

x (2 x - 1) = 1

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