What are the solutions to the equation 20(x^2-2x+1)=x ?

The solutions to the equation 20(x^2-2x+1)=x are x= 5/4 ,x= 4/5

Find the zeros of 20(x^2-2x+1)=x

The zeros of 20(x^2-2x+1)=x are x= 5/4 ,x= 4/5

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

x = \frac{5}{4}, x = \frac{4}{5}

Decimal Notation

x = 1.25, x = 0.8

Solution steps

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

Expand

20 (x^{2} - 2 x + 1) : 20 x^{2} - 40 x + 20

20 x^{2} - 40 x + 20 = x

Move

x

to the left side

20 x^{2} - 41 x + 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 41 ) \pm ( - 41 ) ^{2} - 4 \cdot 20 \cdot 20}{2 \cdot 20}

(- 41)^{2} - 4 \cdot 20 \cdot 20 = 9

x_{1, 2} = \frac{- ( - 41 ) \pm 9}{2 \cdot 20}

Separate the solutions

x_{1} = \frac{- ( - 41 ) + 9}{2 \cdot 20}, x_{2} = \frac{- ( - 41 ) - 9}{2 \cdot 20}

x = \frac{- ( - 41 ) + 9}{2 \cdot 20} : \frac{5}{4}

x = \frac{- ( - 41 ) - 9}{2 \cdot 20} : \frac{4}{5}

The solutions to the quadratic equation are:

x = \frac{5}{4}, x = \frac{4}{5}

x, z = 36 - 3 x^{2} - 3 y^{2}

(x - 8) (3 x + 5) = 0

b = x^{2}

- 45 x^{2} + 250 x + 250 = 0

x, y = x^{2} + 2 x + 4

What are the solutions to the equation 20(x^2-2x+1)=x ?
The solutions to the equation 20(x^2-2x+1)=x are x= 5/4 ,x= 4/5
Find the zeros of 20(x^2-2x+1)=x
The zeros of 20(x^2-2x+1)=x are x= 5/4 ,x= 4/5