What are the solutions to the equation (y(1-y))-2=0 ?

The solutions to the equation (y(1-y))-2=0 are y= 1/2-i(sqrt(7))/2 ,y= 1/2+i(sqrt(7))/2

Find the zeros of (y(1-y))-2=0

The zeros of (y(1-y))-2=0 are y= 1/2-i(sqrt(7))/2 ,y= 1/2+i(sqrt(7))/2

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(y (1 - y)) - 2 = 0

y = \frac{1}{2} - i \frac{7}{2}, y = \frac{1}{2} + i \frac{7}{2}

Solution steps

(y (1 - y)) - 2 = 0

Expand

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

y - y^{2} - 2 = 0

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Simplify

1^{2} - 4 (- 1) (- 2) : 7 i

y_{1, 2} = \frac{- 1 \pm 7 i}{2 ( - 1 )}

Separate the solutions

y_{1} = \frac{- 1 + 7 i}{2 ( - 1 )}, y_{2} = \frac{- 1 - 7 i}{2 ( - 1 )}

y = \frac{- 1 + 7 i}{2 ( - 1 )} : \frac{1}{2} - i \frac{7}{2}

y = \frac{- 1 - 7 i}{2 ( - 1 )} : \frac{1}{2} + i \frac{7}{2}

The solutions to the quadratic equation are:

y = \frac{1}{2} - i \frac{7}{2}, y = \frac{1}{2} + i \frac{7}{2}

18 x^{3} - 2 x^{2} + 27 x - 3

60 = - 0.3 t^{2} + 9.7 t - 2

z^{2} (2 x - y) + 16 (y - 2 x)

ab c + ab c^{2}

k^{3} + 27

What are the solutions to the equation (y(1-y))-2=0 ?
The solutions to the equation (y(1-y))-2=0 are y= 1/2-i(sqrt(7))/2 ,y= 1/2+i(sqrt(7))/2
Find the zeros of (y(1-y))-2=0
The zeros of (y(1-y))-2=0 are y= 1/2-i(sqrt(7))/2 ,y= 1/2+i(sqrt(7))/2