What are the solutions to the equation 1=(1-y)(2-y) ?

The solutions to the equation 1=(1-y)(2-y) are y=(3+sqrt(5))/2 ,y=(3-sqrt(5))/2

Find the zeros of 1=(1-y)(2-y)

The zeros of 1=(1-y)(2-y) are y=(3+sqrt(5))/2 ,y=(3-sqrt(5))/2

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1 = (1 - y) (2 - y)

y = \frac{3 + 5}{2}, y = \frac{3 - 5}{2}

Decimal Notation

y = 2.61803 \dots, y = 0.38196 \dots

Solution steps

1 = (1 - y) (2 - y)

Expand

(1 - y) (2 - y) : 2 - 3 y + y^{2}

1 = 2 - 3 y + y^{2}

Switch sides

2 - 3 y + y^{2} = 1

Move

1

to the left side

y^{2} - 3 y + 1 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot 1 = 5

y_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, y_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

y = \frac{- ( - 3 ) + 5}{2 \cdot 1} : \frac{3 + 5}{2}

y = \frac{- ( - 3 ) - 5}{2 \cdot 1} : \frac{3 - 5}{2}

The solutions to the quadratic equation are:

y = \frac{3 + 5}{2}, y = \frac{3 - 5}{2}

3 x^{2} + 5 = 32

x, 9 x^{2} - k x + 1 = 0

4 b^{2} + 2 b - 2 = 0

10 x^{2} - 12 x = 0

25 x^{2} + 100 x + 100 = 0

What are the solutions to the equation 1=(1-y)(2-y) ?
The solutions to the equation 1=(1-y)(2-y) are y=(3+sqrt(5))/2 ,y=(3-sqrt(5))/2
Find the zeros of 1=(1-y)(2-y)
The zeros of 1=(1-y)(2-y) are y=(3+sqrt(5))/2 ,y=(3-sqrt(5))/2